Questions you marked Missed come first, then Partly (up to 20 at a time). Answer them again, compare, and mark them. Each one you now get right drops off the list.
int x = 1;
switch (x) {
case 1: cout << "A";
case 2: cout << "B"; break;
case 3: cout << "C";
}It prints AB. x is 1, so control enters case 1 and prints A. There is no break after case 1, so execution falls through into case 2, prints B, and only then hits break. case 3 never runs. To print only A, add a break after case 1.
do-while loop instead of a while loop?Use a do-while when the body must run at least once before the condition is checked — because a do-while evaluates its condition after running the body, whereas a while checks first and may run zero times.
do {
x = readInput();
} while (!valid(x)); // ask once, then keep asking until validx after the if?int x = 0;
if (x = 5) {
cout << "entered";
}
// what prints? what is x now?It compiles and runs fine — this is not a compiler error, and calling it a 'bug' would be wrong: it's valid C++. = is assignment, not comparison, so x = 5 stores 5 in x, and the value of that expression is 5, which is truthy — the if body always runs and x is 5 afterward. It's usually a typo for == (comparison), so it's a logic pitfall rather than something the compiler rejects. Many compilers warn about it, and writing if (5 == x) would turn the typo into a real compile error.
switch fall-through intentional rather than a mistake?When several cases should share the same code — you deliberately leave out break so one case flows into the next:
switch (grade) {
case 'A':
case 'B': cout << "Pass"; break; // A and B both -> Pass
case 'F': cout << "Fail"; break;
}int i = 0;
do { std::cout << i; } while (i++ < 2);do-while runs the body first, then checks. Prints 0; 0 < 2 is true (i becomes 1). Prints 1; 1 < 2 is true (i becomes 2). Prints 2; 2 < 2 is false, so the loop ends.
while loop always runs its body at least once.False. while checks the condition before the first run, so it can run zero times. do-while is the loop that always runs at least once.
*p and &x?&x means 'address of x' — it produces the memory address where x lives. *p means 'dereference p' — go to the address stored in p and read/write the value there. They are inverses: *(&x) is just x.
int x = 42;
int* p = &x; // p holds x's address
int y = *p; // y = 42 (value at that address)
*p = 99; // x is now 99p, what does printing p give you versus printing &p?int* p = new int(5);
cout << p; // the heap address
cout << &p; // where p itself lives (stack)p is the value stored in the pointer — the address it points to (here, the heap block from new). &p is the address of the pointer variable itself, which sits on the stack. They are two different addresses. To inspect the heap allocation you print p; &p only tells you where the pointer lives.
A null pointer holds the address 0 — it deliberately points to nothing, and you can safely test for it (if (p) ...). A dangling pointer holds an address that used to be valid but whose memory has been freed or gone out of scope; it still looks like a normal non-zero address, so you cannot detect it by checking — using it is undefined behaviour.
int* p = nullptr; // null: safe to check
int* q = new int(3);
delete q; // q now dangles (still non-null!)Yes. Unlike a reference, a pointer can be reseated any number of times, and can be set to nullptr.
int a=1, b=2;
int* p = &a; // points to a
p = &b; // now points to b
p = nullptr; // now points to nothingThe pointer variable — the 8 bytes holding the address on a 64-bit machine — lives wherever it's declared, typically on the stack as a local. That is separate from what it points to, which can live anywhere: the stack (int* p = &local;), the heap (int* p = new int;), or static storage.
int x = 5; int* p = &x; *p = 9; what is the value of x?9. p holds the address of x, and *p = 9 writes through that address, so it changes x itself.
int a = 1, b = 2;
int* p = &a;
*p = 10;
p = &b;
*p = 20;
std::cout << a << " " << b;*p = 10 writes through p into a. Then p = &b reseats the pointer (a reference could not do that), so *p = 20 changes b, not a.
Stack memory is already committed to the process, so allocating a local is just moving the stack-pointer register down by the needed bytes; freeing is moving it back when the function returns — no searching, no bookkeeping. Heap allocation (new) must ask the allocator to search a free list for a suitable block, mark it used, and return it (and you must delete it later). That search + metadata make it slower.
On typical machines the stack grows downward, toward lower addresses — each call/push decrements the stack pointer. So the 'top of stack' (the most recent item) is actually at the lowest address currently in use, the opposite of the everyday picture of a stack growing up.
int* bad() {
int x = 42;
return &x; // returning the address of a local
}It's a dangling pointer, so dereferencing it is undefined behaviour. x is a local — it lives on bad()'s stack frame; when bad() returns, that frame is reclaimed and x no longer exists, yet the returned pointer still holds its old address. Reading through it may give garbage, may seem to work, or may crash. To return something that outlives the call, allocate on the heap (new) or return by value.
new[] memory with plain delete (instead of delete[])?Undefined behaviour. Array new[] and scalar new can use different bookkeeping, so they must be matched exactly: new ↔ delete, and new[] ↔ delete[]. Mismatching them can corrupt the heap or leak — even if it seems to work on one compiler.
int* a = new int[10];
delete a; // WRONG — must be delete[] a;Each function call pushes a stack frame (return address, saved registers, locals). The stack is small — typically only 1–8 MB. Deep or unbounded recursion keeps pushing frames until the stack is exhausted → stack overflow (a crash). This is one reason large data lives on the heap instead.
Because of ASLR (Address Space Layout Randomization): the OS deliberately randomizes where memory regions are placed each run, so attackers can't predict addresses. The structure is the same but the absolute base shifts, so new int prints a different address each time.
void f() { int a = 1; int* p = new int(2); }, what lives on the heap?Only the int that p points to (created by new). a and the pointer variable p itself are locals on the stack. (And this function leaks that int: nothing deletes it.)
True. A push decrements the stack pointer, so the "top" of the stack is its lowest address.
A reference cannot be null and cannot be reseated (rebound to a different variable after initialization). It also has no arithmetic and must be initialized when declared. A pointer can be null, can point at different things over time, and supports pointer arithmetic.
void byVal(int x){ x = 99; }
void byRef(int& x){ x = 99; }
int n = 1;
byVal(n); byRef(n);
cout << n;It prints 99. byVal gets a copy, so its x = 99 changes only the local copy — n is untouched by it. byRef gets a reference (alias) to n, so its x = 99 writes straight through to n. After both calls, n is 99.
const T& instead of by value?Passing by value (T) copies the whole object, which is expensive for a big std::string or std::vector. A const T& binds an alias — no copy — while const guarantees the function won't modify the caller's object. So you get a reference's speed with read-only safety.
void print(const std::string& s); // no copy, cannot change sBecause a reference is a permanent alias with no identity of its own — it must be bound to an existing object the moment it's created, and can never be re-bound afterward. int& r; is a compile error; int& r = x; is required. (That's exactly why a reference makes a good fixed alias but a poor 'optional' or reseatable handle — use a pointer for those.)
void f(int a, int& b) { a = 10; b = 20; }
int main() {
int x = 1, y = 2;
f(x, y);
std::cout << x << " " << y;
}a is a copy, so a = 10 changes only the copy and x stays 1. b is a reference to y, so b = 20 writes straight into y.
False. A reference must be bound to an existing object when it is declared, and it cannot be reseated, so there is no null reference. Use a pointer when "nothing" is a valid value.
private gives no runtime security, what is the real point of making data private?Two real benefits: (1) Invariant safety — every change must go through a public method that can validate it, so the object can never be put into an illegal state. (2) Freedom to change internals — since outside code can't touch the fields directly, you can later change how the data is stored without breaking any user of the class. private is a design/compile-time tool for controlled change, not a runtime guard against attackers.
An invariant is a rule that must always be true for a valid object — e.g. a BankAccount's balance must never be negative. By keeping balance private and only changing it through deposit()/withdraw(), each of which checks the rule, the invariant is guaranteed after every operation, so the object is never left inconsistent.
A method can validate before writing; a public field can be set to anything:
class Account {
double balance = 0;
public:
bool withdraw(double amt){
if (amt < 0 || amt > balance) return false; // guard
balance -= amt; return true;
}
};If balance were public, any code could do a.balance = -500; and break the invariant. The method is the single validated gate.
class Account {
int balance = 0;
public:
void deposit(int a) { if (a > 0) balance += a; }
bool withdraw(int a) { if (a > balance) return false; balance -= a; return true; }
int get() const { return balance; }
};
int main() {
Account acc;
acc.deposit(100);
acc.withdraw(150);
acc.withdraw(30);
std::cout << acc.get();
}Every change goes through a method that checks it. deposit(100) sets 100; withdraw(150) is refused (it would overdraw), so the balance stays 100; withdraw(30) leaves 70.
private stops other code from reading that object's memory at runtime.False. private is a compile-time rule: the compiler refuses direct access from other code. Nothing protects the bytes at runtime. Its purpose is controlled change and protecting invariants.
class and a struct?(1) Default member access: class members are private by default, struct members are public by default. (2) Default inheritance mode: class B : A is private inheritance by default, struct B : A is public by default. Everything else — constructors, methods, virtual functions, templates, inheritance — is identical.
x accessible from outside here?class C { int x; };
int main(){ C c; c.x = 5; }No. In a class, members are private by default, and x has no access specifier, so it's private — c.x = 5; from outside is a compile error. Change class to struct and x becomes public, so it would compile. That default is the whole practical difference.
struct rather than a class?Use a struct for a passive bundle of public data with little or no behaviour (a plain-old-data aggregate, e.g. struct Point{int x,y;};). Use a class for an encapsulated type that maintains invariants and exposes behaviour through methods while hiding its data. The compiler treats them the same — this is purely a readability convention.
class and struct? Select all that apply.Only the two defaults differ: members are private in a class and public in a struct, and inheritance is private vs public by default. Both can have constructors, virtual functions and everything else.
struct B { int x; }; and class D : B {};. Does D d; d.x = 1; compile?No. A class inherits privately by default, so x becomes private inside D. Writing class D : public B (or struct D : B) makes it compile.
// entity.h
class Entity { int x; public: int GetX(); };
// entity.cpp
int GetX() { return x; } // note: no Entity::The out-of-class definition is missing the Entity:: qualifier, so int GetX(){...} defines an unrelated free function named GetX, not the member. The class's declared GetX() therefore has no definition, and the linker complains. Fix: int Entity::GetX() { return x; }.
const?Mark a method const whenever it doesn't modify the object's state — typically getters: int GetX() const;. The payoff: it can then be called on const objects and const references, and the compiler enforces the read-only promise (modifying a member inside a const method is an error unless the member is mutable). Missing const on getters is something interviewers notice.
You get a duplicate-symbol link error, because of the One Definition Rule (ODR): a non-inline function may be defined only once in the whole program. A header is textually pasted into every .cpp that #includes it, so a full body in the header becomes a definition in each including .cpp → multiple definitions. Fix: put the declaration in the header and the single definition in one .cpp (or mark it inline).
A declaration tells the compiler something exists and its type/signature (int GetX();) — it can appear many times. A definition actually provides the body or storage (int Entity::GetX(){return x;}) — it must appear exactly once (ODR). Declarations live in headers; the single definition lives in a source file.
Entity declares int GetX() const;. Which out-of-class definition is correct?int Entity::GetX() const { return x; }: the Entity:: qualifier ties it to the class, and const must match the declaration. Without Entity:: you define an unrelated global function and the member stays undefined (a link error).
inline.True. That is why such a definition can sit in a header included by many .cpp files without breaking the One Definition Rule.
counter() three times in a row print?int counter(){ static int n = 0; return ++n; }It prints 1 2 3. The static local n is initialized to 0 once, on the first call, then retains its value between calls (lifetime = the whole program, scope = the function). So each call increments the same n: 1, 2, 3. Without static, n would reset to 0 each call and print 1 1 1.
static do to a variable declared at file/global scope?It gives the variable internal linkage — the symbol is visible only within that translation unit (that .cpp). Other files can't see or link to it, and two files can each have their own static int x; with no clash. This is the opposite of extern, which shares one symbol across files.
static local variable initialize safely regardless of other globals' init order?This refers to the 'static initialization order fiasco' — a named C++ problem where globals defined in different .cpp files are constructed in an unspecified order across files. The question asks why a function-local static is immune to that ordering problem.
Because a static local initializes on first use — the first time control reaches its declaration — not at program startup. This dodges the static initialization order fiasco: globals in different translation units are initialized in an unspecified order, so one global's constructor using another (from a different .cpp) may run before that other is initialized (a bug). A static local's init order is instead defined by call order, which is deterministic. (Basis of the Meyers singleton.)
extern int x; declare?It's a declaration (not a definition) saying: 'an int x with external linkage is defined in some other translation unit — share that one.' It reserves no storage itself. If no .cpp actually defines int x; anywhere, the linker can't find the symbol and reports 'undefined reference to x'.
void f() {
static int s = 0;
int n = 0;
++s; ++n;
std::cout << s << n << " ";
}
int main() { f(); f(); f(); }s is static: created once, it keeps its value, so it counts 1, 2, 3. n is a normal local, re-created as 0 on every call, so it is always 1.
a.cpp has static int helper = 5; at file scope. b.cpp says extern int helper; and uses it. What happens?The link fails (undefined reference to 'helper'). File-scope static gives internal linkage: the name is private to a.cpp, so b.cpp's extern finds no definition.
static data member increase each object's sizeof?No. A static member is shared by all objects — there is exactly one copy in static storage, not one per object. So it doesn't live inside the object and doesn't count toward sizeof(obj). (An ordinary non-static member does live in each object and adds to its size.) A classic use is a counter of how many instances exist.
struct Entity { static int count; };
// no definition anywhere
int main(){ Entity::count++; }A non-inline static data member is only declared inside the class — it still needs exactly one definition outside it. Without that, the linker can't find storage for Entity::count. Fix (pre-C++17): add int Entity::count = 0; in one .cpp. Fix (C++17+): declare it inline static int count = 0; inside the class, which provides the definition automatically.
static member function read a normal (non-static) data member?A static member function has no this pointer — it isn't tied to any particular object, and you can call it as Class::fn() with no instance at all. Since per-object data lives in an object and there's no object in hand, there's nothing to read it from. A static function can only touch static members (or data passed in as arguments).
struct Entity {
static int count;
int id;
Entity() : id(++count) {}
};
int Entity::count = 0;
int main() {
Entity a, b, c;
std::cout << Entity::count << " " << b.id << " " << sizeof(Entity);
}One shared count for all objects: three constructions make it 3, and b was second, so b.id is 2. A static member lives outside the objects, so sizeof(Entity) is just the int id: 4.
static member function can read the object's non-static data members.False. A static member function has no this, so there is no object to read from; it can only use static members (or an object passed in). g++ rejects it.
enum E { A = 3, B, C };, what are the values of B and C?B == 4 and C == 5. Enumerators without an explicit value take the previous value + 1. A is set to 3, so B becomes 4 and C becomes 5. (If nothing were assigned at all, they would start at 0.)
enum that enum class fixes.(1) Name leakage: a plain enum's names go into the enclosing scope, so enum Color{Red} and enum Fruit{Red} collide. enum class Color{Red} scopes it as Color::Red. (2) Implicit int conversion: a plain enum converts to int silently, so you can accidentally compare unrelated enums or do arithmetic on them; enum class blocks that (you must static_cast<int>(x) deliberately).
Give it an explicit underlying type:
enum class Color : uint8_t { Red, Green, Blue };The : uint8_t fixes storage to 1 byte — useful when packing many enum values into structs or over the wire.
enum class value to an int?With an explicit cast: int n = static_cast<int>(Color::Green);. It's deliberately not automatic — enum class is type-safe and refuses implicit conversion to int, which prevents accidental mixing of unrelated enums and silent arithmetic bugs. You opt in with the cast only when you really mean it.
enum Level { Low, Mid = 5, High };
std::cout << Low << " " << Mid << " " << High;Values start at 0 and count up from the last assigned one: Low is 0, Mid is set to 5, so High is 6. A plain enum converts to int implicitly, so cout prints the numbers.
enum class Color { Red, Green }; What does int x = Color::Green; do?It doesn't compile. enum class has no implicit conversion to int; write int x = static_cast<int>(Color::Green); (which gives 1).
Entity e; stop compiling once you add a constructor that takes arguments?class Entity {
public:
Entity(int x) { /*...*/ } // now Entity e; fails
};As soon as you declare any constructor of your own, the compiler stops providing the free default (no-argument) constructor. So Entity e; (which needs a no-arg ctor) no longer compiles. Fixes: add your own no-arg ctor, or explicitly request the compiler's with Entity() = default;.
The : member(value), ... list (which initializes members directly, before the body) is required for: (1) const members — they must be initialized, never assigned; and (2) reference members — same reason (a reference must bind at creation). It's also required for member objects whose type has no default constructor.
class A { const int id; int& ref;
A(int i, int& r) : id(i), ref(r) {} // must use init list
};The init list constructs each member once, directly with the right value. Assigning in the body first default-constructs the member, then overwrites it — two steps, wasteful for non-trivial types, and outright illegal for const/reference members. So the init list is both faster and more capable.
Entity e(); actually declare?A function named e that takes no parameters and returns an Entity — not an object. This is the 'most vexing parse': the compiler prefers to read it as a declaration. To create an object, write Entity e; or Entity e{}; (the brace form avoids the ambiguity).
Entity object named e?Entity e; creates the object. Entity e(); is the most vexing parse: it declares a function named e that returns an Entity. The other two do not compile.
A const member and a reference member can't be assigned after construction, and a member object with no default constructor can't be built without arguments, so all three need the initializer list. A std::string has a default constructor, so it can be assigned in the body (the list is still preferred).
Base has a non-virtual destructor — what happens when delete b; runs?struct Base { ~Base(){} }; // non-virtual!
struct Derived : Base { /* owns memory */ };
Base* b = new Derived();
delete b;Only ~Base() runs. delete b goes through a Base* whose destructor is non-virtual, so it uses static dispatch and ~Derived() is never called — anything Derived owns is leaked (the standard calls this undefined behaviour). To make it correct, declare the base destructor virtual (virtual ~Base();); then delete b runs ~Derived() first, then ~Base().
RAII (Resource Acquisition Is Initialization) ties a resource's lifetime to an object: acquire it in the constructor, release it in the destructor. A local object's destructor runs automatically when it leaves scope — including when an exception unwinds the stack — so the resource is always released, with no manual free/close/unlock to forget. std::unique_ptr, std::lock_guard, and std::fstream all rely on this.
If a class needs a custom destructor, it almost certainly also needs a custom copy constructor and copy assignment operator. Why: a class that owns a raw resource and defines only a destructor gets a compiler-generated shallow copy — two objects end up owning the same pointer, and when both destruct, the memory is freed twice (double-free). Defining all three (deep copy) prevents it. (C++11 adds move ctor/assignment → Rule of Five.)
Members are destroyed in the reverse order of their declaration. For an inheritance chain the derived destructor runs first, then the base destructor — the mirror image of construction (base→derived to construct, derived→base to destroy).
struct T {
char name;
T(char n) : name(n) { std::cout << "+" << name; }
~T() { std::cout << "-" << name; }
};
int main() {
T a('A');
{
T b('B');
}
T c('C');
}Objects are destroyed at the end of their scope. B lives only in the inner block, so it dies before C is made. At the end of main, locals die in reverse order of construction: C, then A.
virtual.False. It is undefined behaviour: typically only the base destructor runs, so the derived part is never cleaned up (a leak, or worse). A polymorphic base needs a virtual destructor.
Slicing happens when you copy a derived object into a base object by value — only the base part is copied and the derived part is 'sliced off', so overrides and extra data are lost.
Derived d;
Base b = d; // SLICED: b holds only the Base part
Base& r = d; // OK: alias, full object, polymorphism works
Base* p = &d; // OK: pointer, full objectAvoid it by handling polymorphic objects through base pointers or references, never by value.
protected member?A protected member is accessible to the class's own methods and to the methods of its derived classes ('the family') — but not to outside code. That's the key difference from private, which is accessible only to the class's own methods, not even derived classes. protected lets you share implementation down the inheritance chain while hiding it from the outside world.
Construction: base first, then derived — the base subobject must be fully built before the derived part relies on it. Destruction: the reverse — derived first, then base. A derived object physically contains the base subobject laid out first.
is-a means one type is a kind of another (a Car is a Vehicle) — modelled with inheritance. has-a means one type contains another (a Car has an Engine) — modelled with composition (a member object). A common guideline is to prefer composition unless the relationship is genuinely is-a.
struct Base {
Base() { std::cout << "B"; }
~Base() { std::cout << "~B"; }
};
struct Derived : Base {
Derived() { std::cout << "D"; }
~Derived() { std::cout << "~D"; }
};
int main() { Derived d; }Construction goes base first, then derived (B then D). Destruction is the reverse: derived first, then base (~D then ~B).
struct Animal { virtual std::string name() const { return "animal"; } };
struct Dog : Animal { std::string name() const override { return "dog"; } };
void byValue(Animal a) { std::cout << a.name(); }
void byRef(const Animal& a) { std::cout << a.name(); }
int main() {
Dog d;
byValue(d);
std::cout << " ";
byRef(d);
}byValue takes an Animal by value, so the Dog is sliced: only its Animal part is copied and name() returns animal. byRef keeps the real object, so the virtual call reaches Dog::name.
a->speak() print here, and why?struct Animal { void speak(){cout<<"Animal";} }; // not virtual
struct Dog : Animal { void speak(){cout<<"Dog";} };
Animal* a = new Dog();
a->speak();It prints "Animal". Because speak() is not virtual, the call uses static dispatch — the version is chosen at compile time from the pointer's type (Animal*), so Animal::speak runs regardless of the real object. If you mark speak() virtual, it switches to dynamic dispatch — resolved at runtime from the object's actual type via the vtable — and it would print "Dog" instead.
Each class with virtual functions has one vtable — an array of function pointers, one slot per virtual function (overrides point to the derived versions). Each object stores a hidden vptr pointing to its class's vtable. A call p->speak() follows the object's vptr to its vtable, indexes the fixed slot for speak, and calls whatever pointer is there — so a Dog object's vptr leads to Dog::speak. That runtime lookup is dynamic dispatch.
Memory: every polymorphic object stores a hidden vptr (~8 bytes on 64-bit), plus one vtable per class. Speed: each virtual call adds one indirection (load vptr → load slot → call) and usually can't be inlined, since the target is chosen at runtime. A class with no virtual functions has no vtable/vptr and pays nothing.
override specifier do, and why is it worth adding?override tells the compiler 'this function is meant to override a base virtual with a matching signature.' If it doesn't actually match (a typo, wrong const, wrong parameters), you get a compile error instead of silently creating a brand-new function that never gets called. So it catches a common, hard-to-spot bug and documents intent.
final do in C++?On a virtual function, final forbids any further overriding of it in subclasses. On a class, final forbids inheriting from that class at all. It's a way to lock down an inheritance hierarchy (and can help the compiler devirtualize calls for speed).
struct Base {
void a() { std::cout << "Ba "; }
virtual void b() { std::cout << "Bb "; }
};
struct Der : Base {
void a() { std::cout << "Da "; }
void b() override { std::cout << "Db "; }
};
int main() {
Der d;
Base* p = &d;
p->a();
p->b();
}a() is not virtual, so p->a() uses the pointer type (Base) and prints Ba. b() is virtual, so p->b() uses the actual object (Der) and prints Db.
Nothing. Only classes with at least one virtual function get a vtable and a per-object vptr. (On x86-64, struct { int x; } is 4 bytes; adding one virtual function makes it 16: an 8-byte vptr plus padding.)
A class becomes abstract as soon as it declares at least one pure virtual function (virtual void f() = 0;). An abstract class cannot be instantiated directly — Shape s; is a compile error.
You use pointers or references to it — that's the whole point of an abstract base: Shape* s = new Circle();. As long as the derived class (Circle) overrides every pure virtual, it's concrete and can be created, and you manipulate it polymorphically through the abstract base pointer/reference.
Both rely on pure virtuals, but: an abstract class has at least one pure virtual and may also have data members and implemented (normal virtual) methods. An interface is the special case where every function is pure virtual and there's no data — a pure contract. C++ has no interface keyword; it's a convention built from an all-pure-virtual abstract class.
Because clients hold and delete derived objects through base pointers (Base* p = new Derived; ... delete p;). If the base destructor isn't virtual, delete p only runs ~Base() and the derived part leaks (undefined behaviour). A virtual ~Base() = default; ensures the full derived destructor chain runs.
Yes. An abstract class can mix pure virtuals (which every subclass must implement) with ordinary virtual methods that have bodies (shared default behaviour). This is the Template Method pattern: the base defines a skeleton, forcing subclasses to fill in the pure-virtual steps while reusing the implemented ones.
struct Shape { virtual double area() const = 0; };, which line does not compile?Shape s;: a class with a pure virtual function is abstract and can't be instantiated. Pointers and references to it are fine, and so is a derived class that overrides area().
True. It stays abstract until every pure virtual function has an override, so you still can't create an object of it.
int n = getInput();
const int a = n;
constexpr int b = n;
constexpr int c = 42;a and c compile; b is an error. const int a = n; is fine — const only means 'won't change after init', and initializing it from a runtime value is legal. constexpr int b = n; fails because constexpr demands a value known at compile time, but n came from runtime input. constexpr int c = 42; is fine — 42 is a compile-time constant.
const and constexpr?const means read-only after initialization — but the value may be computed at runtime. constexpr is stronger: the value (or function result) must be computable at compile time, so it can be used where the language requires a compile-time constant (array sizes, template arguments, switch cases). Every constexpr is implicitly const; not every const is constexpr.
volatile a correct tool for synchronizing a variable between threads?No. volatile only tells the compiler 'don't optimize away or cache accesses — re-read from memory each time', which is for memory-mapped hardware registers and signal handlers. It provides no atomicity and no memory-ordering guarantees, so it can't safely synchronize threads. Use std::atomic (or a mutex) for that — those give the real guarantees.
constexpr function run at runtime instead of compile time?When you call it with arguments that aren't known at compile time. Given compile-time constant arguments in a constant context it runs at compile time; given a runtime value it simply runs at runtime like an ordinary function.
constexpr int sq(int x){ return x*x; }
constexpr int a = sq(5); // compile time
int y = read(); int b = sq(y); // runtimen is read at runtime (int n = getInput();). Which of these lines compile? Select all that apply.const only means read-only, so it can take a runtime value. constexpr needs a value the compiler knows, so constexpr int b = n; is an error. 42 is a constant, and c * 2 is too because c is constexpr.
volatile makes ++counter safe to call from two threads.False. volatile only stops the compiler from optimizing the reads and writes away. ++counter is still a read-modify-write that two threads can interleave. Use std::atomic<int> or a mutex.
constexpr int s1 = 5*4; int a[s1];
const int s2 = 5*4; int b[s2];
int s3 = 5*4; int c[s3];The third — int s = 5*4; int c[s]; — is not standard C++. Because s is a plain (non-const) int, it's a runtime value, so c[s] is a VLA (Variable Length Array). The constexpr and const versions are standard, because both give a compile-time constant size.
int a[n] (runtime n) compiled fine on g++ — how do you check whether it's actually standard C++?Add -pedantic: g++ -std=c++17 -pedantic t.cpp warns 'ISO C++ forbids variable length array' — proving it's a non-standard GNU extension, not standard C++. Add -pedantic-errors to make it a hard error. (MSVC rejects it outright.) A clean default compile on g++ only means g++ allows the extension, not that the standard does.
-std=c++11 vs c++20)?No. VLAs were never part of any C++ standard (proposed for C++14, then removed). So the -Wvla warning appears under -pedantic regardless of the -std — it depends on the flag (-pedantic), not the version. It vanishes only when you drop -pedantic (extensions on).
A size that is a compile-time constant: either a constexpr value, or a const integer initialized with a constant expression (e.g. const int n = 4*5;). A plain mutable int — even one obviously set to a fixed value — does not qualify and makes the array a (non-standard) VLA.
sizeof. For a normal array, sizeof(arr) is a compile-time constant. For a VLA, the size isn't known until the program runs, so sizeof(arr) is computed at runtime. If sizeof has to be evaluated at run time, you're looking at a VLA.
int a[n]; with a runtime n compiles silently on g++. Which flag makes g++ warn that it is not standard C++?-pedantic (or -pedantic-errors to make it an error). -Wall, -Wextra, -O2 and picking a newer -std= do not mention the VLA, because g++ allows it as an extension.
-std=c++11 to -std=c++20 makes g++ reject int a[n]; with a runtime n.False. The language version does not change it: g++ accepts the VLA as an extension in every mode. Only -pedantic / -pedantic-errors report it.
In the process's virtual address space. Indexing computes base + i*sizeof(element) and relies on elements sitting one after another virtually. In physical RAM the array's pages can be scattered across arbitrary frames — that's fine, because the page table maps the contiguous virtual pages onto those scattered frames, so the program only ever sees one clean contiguous block.
arr[i]?As base_address + i * sizeof(element) — a single multiply-and-add, no lookup table (which is why indexing is O(1)). This formula is exactly why an array must be contiguous: element i has to sit precisely i element-widths after the start.
int arr[100000] is about 400 KB — should it live on the stack or the heap, and why?On the heap — new int[100000] or, better, std::vector<int>. A local array that big can overflow the stack, which is typically only 1–8 MB and shared by all frames in the call chain. Small fixed arrays are fine on the stack; large ones belong on the heap.
new array allocated immediately?No. The virtual address range is reserved right away (so the array looks allocated and contiguous), but the physical frames are typically handed out lazily, on first touch of each page (via page faults). So memory you allocate but never write may never consume physical RAM.
int arr[5] = {10, 20, 30, 40, 50};
int* p = arr;
std::cout << *(p + 3) << " " << (&arr[4] - &arr[1]);p + 3 points at arr[3], so *(p + 3) is 40. Subtracting two element pointers gives the distance in elements, not bytes: &arr[4] - &arr[1] is 3.
In virtual address space. arr[i] is computed from virtual addresses, so they must be continuous. The physical frames behind the pages can be scattered; the page table maps them.
500 (0x000001F4) laid out byte by byte in memory?As F4 01 00 00 — least-significant byte first. The value's bytes are 00 00 01 F4 (most-significant first, the way we write it); little-endian stores them reversed, so to reconstruct the value you flip them back: F4 01 00 00 → 0x000001F4 = 500. Small values hide this (5 is 05 00 00 00, and the leading 05 looks right).
cout << somePointer;, do you see the reversed little-endian bytes or the real address?You see the real, human-readable address (most-significant-first, e.g. 0x7ffee3a4b5c8) — not reversed. Printing shows the value, and the CPU has already read the bytes back in the correct order. You only see the reversed little-endian bytes if you deliberately inspect raw memory (walking (unsigned char*)&p, or a hex/debugger view).
Yes. A pointer is just a multi-byte integer (8 bytes on 64-bit) that holds an address, so it's stored in the machine's endianness — little-endian on x86, meaning its bytes appear reversed in a raw memory dump, exactly like any int.
Store 1 in an int and look at its first byte:
int n = 1;
char* c = (char*)&n;
if (*c == 1) puts("little-endian"); // low byte holds the 1
else puts("big-endian"); // low byte is 0On little-endian the least-significant byte (the 01) sits at the lowest address, so *c == 1.
Single-byte types — char, uint8_t, bool — because one byte has no ordering to reverse. Endianness only concerns multi-byte values (short, int, long, pointers), and even then it's visible only when you look at the raw bytes (hex dump, char*, or data sent between machines).
int x = 0x01020304;
unsigned char* b = (unsigned char*)&x;
std::cout << (int)b[0];Little-endian stores the least-significant byte at the lowest address. 0x01020304 is stored as 04 03 02 01, so the first byte b[0] is 4.
std::cout << x; prints the value of x with its bytes reversed.False. cout prints the actual value. Endianness only shows up when you look at the raw bytes in memory (a hex dump or a char* walk).
4 and exits cleanly. Is it correct?std::array<int,5> a;
a[10] = 4;
cout << a[10];No — it is undefined behaviour. a has valid indices 0–4, but you write to index 10. std::array::operator[] does no bounds checking, so it just computes an out-of-bounds address and writes there. It happened not to crash and printed 4, but that's luck — the same code could crash, corrupt another variable, or behave differently on another compiler/run. 'It printed 4' is not proof of correctness.
.at(i) is bounds-checked. On an out-of-range index it throws std::out_of_range (with a descriptive message via e.what()), instead of silently corrupting memory like []. So a.at(10) on a size-5 array throws rather than doing UB.
a.at(10) = 4; // throws std::out_of_rangeoperator[] unchecked by default?Because of the principle 'don't pay for what you don't use' — a bounds check on every element access adds a comparison that's costly in tight numeric loops, so [] is fast and trusts the programmer. (Java/Python bounds-check always: safer, but you pay on every access with no opt-out.) C++ gives you both: fast [] when you've verified the index, and checked .at() when you want the safety net.
It may: (1) appear to work (produce the expected output, as in the a[10] example), (2) crash (e.g. a segfault), or (3) silently corrupt other memory (overwrite a nearby variable, the return address, etc.). Because the behaviour is undefined it can also differ across compilers, optimization levels, and runs — which is what makes UB dangerous.
-fsanitize=address (AddressSanitizer) — it instruments the program and reports a stack-buffer-overflow when the bad access actually runs. Add ,undefined (-fsanitize=address,undefined) to also catch other undefined behaviour with UBSan.
std::array<int, 5> a{}; a[10] = 4; What does the C++ standard say happens?Undefined behaviour. operator[] does no bounds check, so anything can happen: a crash, silent corruption, or apparently working. It still compiles (g++ accepts it), and only .at() would throw.
std::vector<int> v = {1, 2, 3};
try {
v.at(5) = 9;
std::cout << "ok";
} catch (const std::out_of_range&) {
std::cout << "caught";
}.at() is bounds-checked: index 5 is outside a size-3 vector, so it throws std::out_of_range before anything is written, "ok" is skipped, and the catch prints caught.
std::array rather than std::vector?Choose std::array<T,N> when the element count is fixed and known at compile time — it lives on the stack, has zero heap overhead, and never resizes. Choose std::vector<T> instead when the count is unknown or changes at runtime — it lives on the heap and grows via push_back/resize. Rule of thumb: fixed → array, growable → vector.
std::vector grow when you keep adding elements?You add with push_back (or resize/insert); when it runs out of room it allocates a bigger block on the heap, moves/copies the existing elements into it, and frees the old block. Its element data always lives on the heap (the small vector object itself — pointer, size, capacity — may sit on the stack).
size() and its capacity()?size() is the number of elements currently in use. capacity() is how many elements the currently-allocated block can hold before it must reallocate. Capacity is usually ≥ size and grows in chunks (often doubling), so not every push_back triggers a reallocation. std::array has only a fixed size — no separate capacity.
array and vector offer a bounds-checked access?Yes — both provide .at(i) (bounds-checked, throws std::out_of_range) and operator[] (unchecked, fast). The choice between checked and unchecked access is the same for both containers.
push_back?Only std::vector. std::array<int, 5> and int[5] have a size fixed at compile time and no push_back.
std::vector<int> v;
v.reserve(10);
v.push_back(1);
v.push_back(2);
std::cout << v.size() << " " << (v.capacity() >= 10);reserve(10) allocates room for at least 10 elements but adds none, so after two push_back calls size() is 2 while capacity() is still at least 10 (true prints as 1).
a.at(10) on a size-5 array printed 'Aborted (core dumped)'. Whose fault is the crash?The caller's — for not handling the exception. .at() did its job: it detected the out-of-range index and threw std::out_of_range. Because nothing caught it, the exception propagated up to std::terminate, which aborts the program. The crash means 'an error was reported and nobody handled it', not that .at() is broken. (Contrast [], which wouldn't even tell you.)
Wrap it in try/catch:
try {
a.at(10) = 4;
} catch (const std::out_of_range& e) {
std::cout << "Bad index: " << e.what() << '\n';
}
std::cout << "still running\n"; // reachedNow the exception is caught and handled, and execution continues past the try/catch instead of aborting.
const reference, and why?By const reference (catch (const std::exception& e)). Catching by value copies the exception object and can slice it — if the real exception is a derived type, the copy keeps only the base part, losing information (and its overridden what()). A const& binds to the actual object with no copy and no slicing.
e.what() return?A C-style string (const char*) describing the error — the exception's message. For std::out_of_range from .at(), it includes details such as the bad index and the container size.
try {
std::cout << "A";
throw std::runtime_error("boom");
std::cout << "B";
} catch (const std::exception& e) {
std::cout << "C" << e.what();
}
std::cout << "D";A prints, then throw jumps straight to the matching catch, so B never runs. The handler prints C and the message boom, and the program continues to print D.
catch (const std::exception& e). By reference there is no copy and no slicing (a derived exception keeps its real type and message); const because you only read it.
-fsanitize=address and compiling produced no error — why hasn't AddressSanitizer reported anything yet?Because AddressSanitizer is a runtime detector, not a compile-time one. Compiling only instruments the code; the error is reported when the bad access actually executes. You must run the program (./prog) and reach the offending line for ASan to print its report. A clean compile — or a run that never hits the bug — shows nothing.
It surrounds each variable/allocation with poisoned redzones (guard bytes with no real data) and keeps 'shadow memory' marking which bytes are valid. The instant your program reads or writes into a redzone (or freed memory), ASan detects the poisoned access and reports it — with the file/line, the variable, and byte offsets. They're essentially tripwires around your memory.
By default it aborts on the first error (you'll see ==pid==ABORTING). To make it report the error and keep going, set the environment option ASAN_OPTIONS=halt_on_error=0 before running.
UBSan (-fsanitize=undefined) for undefined behaviour like signed overflow or bad shifts; valgrind (memcheck) for leaks and invalid accesses; and gdb for interactive debugging / core-dump inspection. ASan + UBSan together (-fsanitize=address,undefined -g) catch a large class of bugs during testing.
-fsanitize=address -g: the first adds the runtime checks, -g adds debug info so the report names files and lines.
-fsanitize=address compiles without errors, it has no memory bugs.False. ASan is a runtime detector: it reports a bad access only when that code actually runs. A clean compile (or a run that never reaches the bug) proves nothing.
So it doesn't need one big contiguous block of physical RAM. Once the process is cut into equal-size pages, each page can be placed into any free frame anywhere in RAM — no contiguous run required. This eliminates external fragmentation (enough total memory free, but scattered in gaps too small to use) and unlocks virtual memory, isolation, and sharing.
A page is a fixed-size chunk of the program's content (a slice of code/heap/stack) — think of a chapter torn from a book. A frame is a fixed-size slot in physical RAM — a shelf slot that holds one page. They're the same size, and exactly one page fits in one frame. 'Page 3 is in frame 47' means page 3's actual bytes are stored in that physical frame.
No. A virtual address is a per-process, imaginary label — it holds nothing itself and consumes no RAM. The real bytes live in physical frames; the page table translates each virtual address to the frame that actually stores the data. (Two different processes can even print the same virtual address mapping to completely different physical memory.)
The page table maps each contiguous virtual page to whatever physical frame holds it, and the MMU performs that translation on every memory access. So the program always sees a clean, ordered, contiguous virtual layout, while the hardware quietly redirects each access to the scattered real location. Non-contiguous in RAM, contiguous to the program.
(1) Virtual memory — a process can use more memory than physically exists (unused pages live on disk). (2) Isolation/protection — each process has its own page table, so it can't see another process's memory, and pages carry permission bits. (3) Sharing — one physical frame (e.g. a shared library like libc) can be mapped into many processes at once.
False. Pages and frames are the same size, and one frame holds exactly one page.
External fragmentation. A process no longer needs one big contiguous block of RAM, because any free frame can hold any page. (Paging can cause a little internal fragmentation in the last page.)
The MMU (Memory Management Unit) is dedicated hardware in the CPU that translates a virtual address into a physical address on every memory access — by walking the page table (or hitting the TLB) — and simultaneously checks the permission bits (present? writable? executable?), raising a page fault or protection fault if something is wrong. Software sets up the page tables; the MMU reads them and translates automatically.
Page number and offset. The page number selects which page — looked up in the page table to get the physical frame number. The offset selects where inside that page/frame the byte is. Physical address = (frame base) + offset.
12290 with 4KB pages -> page 3 (12290/4096), offset 2
page 3 -> frame 47 ; physical = frame47_base + 2Because it would be enormous. With a 48-bit address and 4 KB pages there are 2^36 pages; a flat table of 8-byte entries is 2^36 × 8 = ~512 GB per process — impossible, and almost all empty since a process uses a tiny slice of its address space. Multi-level paging uses a sparse tree of small (512-entry) tables and creates only the sub-tables for regions actually in use, shrinking it to a few KB.
Four levels. The 48-bit virtual address splits into four 9-bit indexes (one per level, each into a 512-entry table) plus a 12-bit offset (4 KB pages): 9 + 9 + 9 + 9 + 12 = 48. The MMU walks Table1→Table2→Table3→Table4 to reach the frame. (Newer CPUs offer a 5-level mode for a larger 57-bit space.)
The TLB (Translation Lookaside Buffer) — a small, fast cache in the MMU that stores recent page→frame translations. On a TLB hit, the translation is instant and the multi-level walk is skipped. Thanks to locality, most accesses hit the TLB, so the expensive 4-level walk happens only on a miss.
12290 = 3 × 4096 + 2, so it is page 3, offset 2. The page number goes through the page table to find the frame; the offset is kept as-is inside that frame.
12 bits, because 2^12 = 4096 bytes. The remaining bits are the page number (on x86-64: 4 × 9 bits for the four table levels).
External fragmentation wastes memory in the gaps between allocations: enough total memory is free, but it's scattered in pieces too small (or separated by live allocations) to satisfy a large contiguous request. Internal fragmentation wastes memory inside an allocated block — the unused tail of the last page of a region. The names say it: between blocks (external) vs inside a block (internal).
Paging places each page into any free frame, so there are no unusable gaps between allocations → external fragmentation is gone. But a region rarely fills a whole number of pages, so the last page of each region is usually only partly used — the remainder is wasted inside that frame → internal fragmentation (bounded by page size − 1). It trades a large, scattered waste for a small, predictable one.
Because all the earlier pages are completely full of the region's data — they flow continuously with no gaps. Only when you reach the end do you run out of data mid-page, leaving the tail of the final page unused. Every page before it is 100% packed, so there's nothing to waste there.
No — it's a problem / side effect, not a method. The methods are contiguous allocation, paging, and segmentation. Contiguous allocation and segmentation suffer from external fragmentation (variable-size, must-be-contiguous placement); paging is the method that cures it. You never 'choose external fragmentation' — it happens to you when you use a contiguous scheme.
A base (the segment's start address in physical memory) and a limit (its length). An address is (segment number, offset). The limit provides protection: if the offset is ≥ the limit, the access is past the segment's end and is rejected — so a program can't read/write outside its segment. Segmentation is good for logical protection/sharing but reintroduces external fragmentation.
10 / 4 = 2.5, and you can't use half a frame, so it takes 3 frames (12 KB). The last frame holds only 2 KB of data, so 2 KB is wasted inside it.
False. Any free frame can hold any page, so gaps between allocations never block a request. Paging trades that for a little internal fragmentation in the last page.
1) The program touches a virtual page whose page-table entry is marked not present. 2) The CPU raises a page fault — a trap into the OS. 3) The OS finds a free frame (evicting a cold page to disk first if needed), 4) loads the page's bytes from disk into that frame, 5) updates the page table to mark it present and point at the frame, and 6) restarts the faulting instruction. The program resumes as if nothing happened — it just felt a small pause.
Because the cost is paid once per page, then that page sits in RAM for many fast reuses. Thanks to locality of reference, a program hammers a small working set of pages, so faults are rare relative to total accesses. Demand paging also avoids loading (and wasting RAM on) the large parts of a program you never touch, and speeds startup. Many faults are also 'minor' (no disk I/O). It only loses when the working set doesn't fit in RAM.
Thrashing is when the working set is larger than available RAM, so the OS is forced to constantly evict a page and almost immediately fault it back in. The machine spends most of its time doing disk I/O for paging instead of real work, and performance collapses. The fix is more RAM or a smaller working set (fewer concurrent processes).
A minor fault is resolved without disk I/O — the page is already in RAM (in the page cache, shared by another process, or a fresh zero-filled page); the OS just fixes the mapping. A major fault requires reading the page from disk, which is the slow, expensive kind. Most faults after warm-up are minor.
A page fault is a normal event: the address is valid, the page just isn't in RAM yet, so the OS transparently loads it and continues. A segmentation fault is an error: the program accessed memory it isn't allowed to (unmapped or wrong permission), so the OS refuses and usually kills the process. Same hardware mechanism (a trap), opposite outcomes.
Swapping is evicting cold (rarely used) pages from RAM out to a reserved area on disk (the swap/page file) when RAM is full, freeing frames for pages needed now. It's what makes virtual memory possible: total memory in use across processes can exceed physical RAM, because RAM acts as a fast window over a larger space that overflows to disk.
A page fault is normal: the OS loads the missing page and restarts the instruction. A major fault needs a disk read (a minor one doesn't). It is not a bug and not a segfault, which is an illegal access the OS refuses.
Thrashing: the working set is bigger than RAM, so the system spends its time on page faults instead of useful work.
open("config.txt"); it works from one directory but fails with 'file not found' from another. Why?"config.txt" is a relative path — it's resolved against the process's current working directory, not the location of the executable. Run the program from a directory that doesn't contain config.txt and the open fails. Fixes: use an absolute path, compute the path relative to the executable, or chdir to the expected directory. Very common bug.
An absolute path starts at the root / (e.g. /home/anup/config.txt) and means the same thing no matter your current directory. A relative path (e.g. config.txt, ../data/x) is resolved from the current working directory, so its meaning changes with where you run the program. Absolute is the location-independent one.
. and .. refer to in a path?. is the current directory; .. is its parent directory. So ./x is x here, and ../x is x one level up. They're used constantly in relative paths and in cd ...
ls -la show that a plain ls does not?Two things: -l gives the long format (permissions, owner, group, size, timestamp per entry), and -a shows all entries including hidden dotfiles (names starting with ., plus the . and .. entries). Plain ls lists only the visible names.
cd /home/anup/project/src
cd ../build
pwd.. is the parent of src, which is /home/anup/project; then into build.
/etc/hosts: an absolute path starts at the root /. The others are relative to the current directory.
-rwxr-xr-x correspond to?755. Ignore the leading - (it just marks a regular file; d would be a directory). Read the three triplets with r=4, w=2, x=1: owner rwx = 4+2+1 = 7, group r-x = 4+0+1 = 5, others r-x = 5 → 755.
./prog says 'Permission denied' — what's the most likely cause?The binary is missing its execute (x) bit, so the kernel won't run it. Fix it with chmod +x prog (or chmod 755 prog). This is the #1 cause of 'Permission denied' for a script or a freshly created binary. (A compiled binary is usually 755; a source .cpp is 644 — no x, which is correct since you don't execute source.)
The three triplets are for the file's owner, its group, and others. On each access the kernel checks your process's UID/GID: if you're the owner it uses the owner triplet; else if you're in the file's group, the group triplet; else the others triplet. Exactly one triplet applies — chosen most-specific-first — and it decides whether your requested r/w/x is allowed.
On a directory: r = list the names inside it; w = add/remove/rename entries (create or delete files in it); x = enter/traverse it (cd into it and reach files by name). Notably you need x on a directory just to reach files inside it, even if you already know their paths.
-rw-r--r-- in numeric form?Owner rw- = 4+2 = 6, group r-- = 4, others r-- = 4, so 644 (the usual mode for a source file).
ls -l shows drwxr-x---. What is it, and can others list its contents?The leading d means a directory. The last triplet (others) is ---, so others have no r and cannot list it.
chmod 755 f and chmod +x f?chmod 755 f sets all three triplets explicitly and absolutely — owner rwx, group r-x, others r-x — overwriting whatever was there. chmod +x f is relative: it only adds the execute bit (for all three categories) and leaves the read/write bits untouched. Use numeric for an exact mode; symbolic to nudge one bit.
chmod, how do you give execute permission to ONLY the owner?chmod u+x file — u targets the owner (user) and +x adds execute, leaving group and others untouched. The target letters are u=owner, g=group, o=others, a=all. (chmod +x file with no letter behaves like a+x — execute for everyone.)
./script.sh?The execute (x) bit. Without it, the shell won't run the file even if it's readable. Fix with chmod +x script.sh. (For a script you also need a readable interpreter, but the missing x is the classic cause.)
-rw-r--r-- (644). What does ls -l show after chmod u+x f?u+x adds execute for the owner only and leaves every other bit alone: -rwxr--r-- (744).
rwxr-xr-x, whatever its bits were before?chmod 755 f: a numeric mode sets all three triplets at once. The symbolic forms only add or remove the bits you name.
chown?Changing a file's owner requires root. If ordinary users could hand files away, they could dodge disk quotas (dump big files onto someone else) or plant files as another user — a security hole. So only root can change ownership. A regular user can change a file's group to one they belong to (chgrp), but not its owner.
chown user:group file # owner AND group
chown user file # owner only
chown :group file # group only (or: chgrp group file)
chown -R user:group dir # recursive: dir and everything insideThey're checked together on every open/read/write/execute. The kernel compares the process's UID/GID to the file's owner and group to pick which rwx triplet applies (owner, group, or others), then checks whether that triplet grants the requested operation. So 'who owns it' selects which permissions, and the rwx bits say what those permissions are.
id show that whoami does not?whoami prints just your username. id prints your numeric UID, primary GID, and all group memberships. That extra detail is what matters for permissions, because the kernel checks those numeric IDs (not the name) against a file's owner/group to decide which rwx triplet applies to you.
chown.False. Only root can change a file's owner (chown: ... Operation not permitted otherwise). A regular owner can only change the group, and only to a group they belong to.
data.txt to devs?chown :devs data.txt: the empty part before the colon leaves the owner unchanged. (chgrp devs data.txt does the same.) chown devs data.txt would change the owner.
The list initializes a member directly, once, as the object is built. In the body the member already exists (default-constructed / indeterminate), so a = x; only assigns — for class members that's default-construct + reassign (two steps) vs one.
const members, reference members, member objects whose type has no default constructor, and a base class with no default constructor.
In the derived class's initializer list: Derived(int x) : Base(x) {}. It's the only place you can — omit it and the base's default ctor is called (error if none). Needed even inside a default constructor.
In the order they are declared in the class, not the order written in the list (bases first, then members). So don't initialize one member from another that's declared after it.
One constructor calls another of the same class through the list (Entity() : Entity(0) {}) to reuse setup. Rule: the delegation must be the only item in the list.
int a = 0; at the declaration, why is the init list still needed?A default member initializer is a fixed fallback, same for every object, and can't use constructor arguments or initialize a base class. The list is needed for argument-driven values and for base class initialization.
struct Log { Log(char c) { std::cout << c; } };
struct S {
Log x, y;
S() : y('Y'), x('X') {}
};
int main() { S s; }Members are initialized in the order they are declared (x, then y), not the order written in the list. g++ even warns about it with -Wall (-Wreorder).
int a = 0;) and is also in the member initializer list, the list's value is used.True. The list takes precedence: the default initializer is simply not used for that constructor (the member is initialized once, with the list's value).
An expression — it produces a value. That's why it can be used inline where a value is needed (initializing a variable or a const, a return, a function argument), which an if/else statement can't do.
const when if/else can't?A const must get its value at the moment of declaration, in one shot. The ternary is a single expression that yields that value. if/else runs after the variable exists, forcing declare-then-assign — which const forbids.
cond ? 1 : 2.5, what is the result type and why?double. The two branches must have a common type; int and double convert to the common type double, so 1 becomes 1.0 and the whole expression is double — regardless of which branch is chosen.
When they have no common type to convert to — e.g. cond ? 5 : "hi" (an int and a const char*). Same type isn't required; a common/convertible type is.
cout << c ? "a" : "b"; need parentheses?Because << has higher precedence than ?:, so it parses as (cout << c) ? "a" : "b" — it prints c, then applies ?: to the stream (broken). cout << (c ? "a" : "b") forces the ternary first.
No — only the chosen branch is evaluated. That's what makes p ? *p : 0 safe: *p runs only when p is non-null.
int x = 7;
std::cout << (x % 2 == 0 ? "even" : "odd") << " " << (x > 5 ? x * 2 : x);7 is odd, so the first ternary gives "odd". 7 > 5, so the second gives x * 2 = 14. Both are wrapped in parentheses because << binds tighter than ?:.
auto r = true ? 1 : 2.5; What is the type of r?double. The two branches need a common type: int and double meet at double, so even though the 1 branch is chosen, r is the double 1.0.
The stack and the heap. A stack object (Entity e;) has automatic lifetime — destroyed when it leaves scope, with its destructor called for you. A heap object (new Entity) has manual lifetime — it lives until you delete it, so it can outlive the scope (but leaks if you forget).
-> but a stack object with .?new gives you a pointer, not the object itself, so you must go through the pointer: p->x, which is shorthand for (*p).x (dereference, then dot). A stack variable is the object, so you use . directly.
Heap memory that was allocated with new but never deleted. The program still owns the address but nothing uses it and nothing else can reuse it, so memory is wasted until the program exits.
A pointer holding an address whose object no longer exists — using it is undefined behaviour. Two ways: (1) point at a stack object and let it go out of scope; (2) delete a heap object but keep using the pointer (use-after-delete).
nullptr after delete?After delete the pointer still holds the freed address (dangling). Setting it to nullptr marks it as 'points to nothing' so you can test it, and because delete nullptr; is a safe no-op, it also protects against an accidental double-delete.
Entity e(); actually declare?A function named e taking no arguments and returning an Entity — not an object (the 'most vexing parse'). To make an object use Entity e; or Entity e{};.
Entity* e = new Entity; How do you call its print() method?e->print(): e is a pointer, and -> dereferences then accesses the member (the same as (*e).print()). e.print() does not compile on a pointer.
Losing the only pointer to a new object without delete leaks it, and calling through a pointer after delete uses a dangling pointer. Pointing at a live stack object in the same scope is fine, and make_unique cleans up by itself.
new do?It (1) allocates memory on the heap (via operator new, usually malloc underneath) and (2) calls the constructor on that memory. It returns a typed pointer. (delete does the reverse: destructor, then free.)
new differ from malloc? (three ways)new calls the constructor (and delete the destructor); new returns a typed pointer (no cast) while malloc returns void*; and new throws std::bad_alloc on failure while malloc returns nullptr.
new[] be paired with delete[] and not plain delete?new[] stores a count of the elements; delete[] reads it and runs the destructor for every element before freeing. Plain delete runs at most one destructor and mismatches the allocation bookkeeping → undefined behaviour.
delete and free?delete first calls the destructor, then frees the memory (via operator delete, usually free). free only frees the raw memory — no destructor. So delete = destructor + free; free = free only.
free/delete are given only p — how do they know where the header is and how big the block is?The header offset is a fixed constant compiled into the allocator, so it computes p - headerSize to reach the header from just p. The block size is stored inside that header, so it reads it once the header is found.
new int; and new int();?new int allocates the memory but leaves the value indeterminate (garbage). new int() value-initializes it to 0. (For a class with a default constructor, both forms run it — the difference only matters for plain types.)
struct C {
C() { std::cout << "c"; }
~C() { std::cout << "d"; }
};
int main() {
C* p = new C[3];
delete[] p;
}new C[3] allocates the array and runs the constructor for each element (ccc). delete[] runs the destructor for each element (ddd) before freeing. That's why new[] must be paired with delete[].
new compared with malloc? Select all that apply.new calls the constructor and returns a typed pointer (no cast). By default it throws std::bad_alloc when memory runs out; it's malloc that returns nullptr. malloc never calls a constructor.
A constructor callable with a single argument (and not marked explicit). The compiler can use it to implicitly convert that argument's type into the class — e.g. Entity(int) lets Entity e = 22; work.
Entity e = "Anup"; fail?Because it would need two user-defined conversions: const char* -> std::string -> Entity. The compiler allows at most one user-defined conversion in a chain. Entity e = std::string("Anup"); works because that's only one.
explicit keyword do?It marks a constructor (or conversion operator) as usable only for direct/explicit construction, disabling implicit conversion through it. So Entity e = 22; becomes an error while Entity e(22); still works.
explicit constructor, which works: Entity e = 22; or Entity e(22);? Why?Entity e(22); works — that's direct initialization, which explicit allows. Entity e = 22; is copy initialization, which relies on implicit conversion, and that's exactly what explicit blocks.
explicit often a good idea?To stop accidental/surprising implicit conversions that hide bugs (e.g. String s = 65; silently building a 65-char string). explicit forces the caller to state intent; you drop it only when you truly want the implicit conversion.
struct Entity { explicit Entity(int a); }; and void f(Entity);, which lines compile? Select all that apply.explicit blocks implicit conversion: Entity e = 22; (copy-initialization) and f(22) both need it, so they fail. Direct construction Entity e(22); and the explicit Entity(22) still work.
struct Meters {
double v;
Meters(double x) : v(x) {}
};
double twice(Meters m) { return m.v * 2; }
int main() { std::cout << twice(3.5); }Meters(double) is a non-explicit single-argument constructor, so the compiler silently converts 3.5 into a Meters to call twice. 3.5 × 2 = 7 (printed as 7).
It defines what an operator (+, ==, <<...) does for your own type — an operator is just a function with a special name. It's compile-time polymorphism because the compiler chooses which operator function to call from the operand types while compiling (unlike virtual functions, chosen at runtime).
When the left operand is not your class. A member operator's left operand is always this, so for cout << e (left operand is std::ostream) or 2 * v (left operand is int) you can't use a member — you write a free function taking both operands as parameters.
operator<< be a free function and return std::ostream&?Free because its left operand is the stream (std::ostream), which you don't own and can't add members to. It returns the stream by reference so << calls chain: cout << a << b groups as (cout << a) << b, and each << must hand back the stream for the next one.
operator+ return by value but operator= returns *this by reference?a + b produces a brand-new object, so you return it by value (returning a reference to that local would dangle). a = b modifies the existing left object (which is *this), so you return *this; by reference — no copy, and it lets a = b = c chain.
Any two of: you can't invent new operators or change their precedence/arity; at least one operand must be a user-defined type (int+int can't be redefined); and ::, ., .*, ?:, sizeof can't be overloaded at all.
struct V {
int x, y;
V operator+(const V& o) const { return {x + o.x, y + o.y}; }
V& operator+=(const V& o) { x += o.x; y += o.y; return *this; }
};
std::ostream& operator<<(std::ostream& os, const V& v) {
return os << "(" << v.x << "," << v.y << ")";
}
int main() {
V a{1, 2}, b{3, 4};
V c = a + b;
a += b;
a += b;
std::cout << c << " " << a;
}a + b builds a new V (4,6) and leaves a alone. += changes a itself: (1,2) → (4,6) → (7,10). operator<< returns the stream so the calls chain.
:: (scope resolution) can't be overloaded, and neither can ., .*, ?: or sizeof. +, << and [] are all overloadable.
this, and where is it available?this is a pointer holding the address of the current object — the object a member function was called on. It exists inside every non-static member function (static functions don't have it).
this in a non-const vs a const member function?Non-const method: Entity* const — a const pointer to a non-const object (you can modify the object, not reseat the pointer). Const method: const Entity* const — you can't modify the object either. The trailing const after the parameter list makes it a const member function.
this and *this?this is the pointer (the object's address); *this is the object itself (dereferenced). You return *this by reference (Entity&) to hand back the same object for chaining; returning this would return the address (Entity*).
this->?Only to disambiguate a member from a same-named parameter (e.g. this->a = a;), or in some template code (dependent base-class names). Otherwise you can access members directly and this-> is optional.
static member function have no this?Because it isn't tied to any particular object — you call it as Class::fn() with no instance. With no object, there's no address to put in this, which is also why a static function can only touch static members.
struct Counter {
int n = 0;
Counter& add(int k) { n += k; return *this; }
};
int main() {
Counter c;
c.add(1).add(2).add(3);
std::cout << c.n;
}add returns *this by reference, so every call in the chain works on the same object c: 1 + 2 + 3 = 6.
Counter by value instead of by reference. What does it print now?struct Counter {
int n = 0;
Counter add(int k) { n += k; return *this; } // returns a copy
};
int main() {
Counter c;
c.add(1).add(2).add(3);
std::cout << c.n;
}Returning *this by value hands back a copy. Only the first add(1) changes c; .add(2) and .add(3) change temporary copies that are thrown away. So c.n is 1.
Automatic (stack): local variables, alive until the end of their scope. Dynamic (heap): created with new, alive until you delete it. Static/global: static locals, globals, static members — alive for the whole program.
The local lives in the function's stack frame, which is reclaimed when the function returns. The returned address then points to memory that's no longer valid — a dangling pointer — so using it is undefined behaviour.
Any two of: return by value (the caller gets its own copy); allocate on the heap and transfer ownership (return the pointer / a smart pointer); or have the caller pass in a buffer (by reference/pointer) that the function fills.
RAII (Resource Acquisition Is Initialization): acquire a resource in the constructor and release it in the destructor. It's exception-safe because when an exception propagates, the stack unwinds and the destructors of all fully-constructed automatic objects run — so the resource is always released, even on an error path.
static local variable keep its value between calls?Because it has static storage duration (program lifetime) rather than automatic: it's created and initialized once, on first reach, and persists across calls. Only its scope is local; its lifetime is the whole program.
id.)struct T {
int id;
T(int i) : id(i) {}
~T() { std::cout << id; }
};
int main() {
T a(1);
{
T b(2);
static T s(9);
}
T c(3);
}b is automatic and dies at the end of the inner block: 2. At the end of main, the remaining automatics die in reverse order: c then a: 31. The static local s lives until the program exits, so it dies last: 9.
int& f() { int x = 5; return x; }: x dies when f returns, so the reference points at a dead stack slot (g++ warns: reference to local variable 'x' returned). Returning by value, a static local, or a unique_ptr are all safe.
They remove manual new/delete — no leaks, double-frees, or dangling. They work via RAII: the smart pointer's destructor deletes the owned object automatically when the smart pointer goes out of scope.
unique_ptr and shared_ptr, and how does shared_ptr know when to delete?unique_ptr has a single owner (can't be copied, only moved). shared_ptr allows many owners via a reference count: copying bumps it up, destruction bumps it down, and the object is deleted when the count reaches 0.
weak_ptr for, and why doesn't it affect the reference count?It's a non-owning observer of a shared_ptr, used to break circular references (a cycle of shared_ptrs never reaches count 0 -> leak) and to safely check/access the object via .lock(). It doesn't count because it doesn't own the object — it only watches it.
make_unique/make_shared over raw new?They're exception-safe and keep raw new out of your code. make_shared also does a single allocation for the object and its control block (vs two with shared_ptr<T>(new T)).
release(), reset(), and std::move on a unique_ptr?release() gives up ownership without deleting and returns the raw pointer (you must delete it). reset() deletes the owned object now. std::move transfers ownership to another unique_ptr (which will delete it); the source becomes null.
For non-owning references — 'borrowing' an object whose lifetime is managed elsewhere. You use it to read/call but never delete through it, and it's valid only while the real owner keeps the object alive (a reference is often even better when it can't be null).
auto a = std::make_shared<int>(5);
std::cout << a.use_count();
{
auto b = a;
std::cout << a.use_count();
}
std::weak_ptr<int> w = a;
std::cout << a.use_count();One owner: count 1. Inside the block b shares ownership: 2. b dies at the closing brace: back to 1. A weak_ptr observes without owning, so the count stays 1.
auto p = std::make_unique<int>(1); auto q = p; What happens?It does not compile. A unique_ptr is the sole owner, so it can't be copied; transfer ownership with auto q = std::move(p); (then p is empty).
Any two of: initializing a new object from another (T b = a; or T b(a);), passing an object to a function by value, or returning an object by value (though that copy is often elided).
It does a shallow, member-wise copy — copying each member's value. For a pointer that copies the address, so both objects point at the same heap buffer. Then both destructors delete it -> double-free, and changing one affects the other.
A deep copy allocates a separate buffer for the new object and copies the contents into it, so each object owns independent memory. Now each destructor deletes its own buffer — no shared pointer, no double-free.
Rule of Three: if a class needs a custom destructor, it almost certainly also needs a custom copy constructor and copy assignment operator (all manage the same resource). Rule of Five (C++11) adds a move constructor and move assignment for efficiency.
The copy constructor builds a brand-new object from an existing one (T b = a;). Copy assignment (operator=) is used when both objects already exist (b = a;): it must free its own old resource, deep-copy, guard against self-assignment, and return *this.
D = default constructor, C = copy constructor, A = copy assignment.)struct S {
S() { std::cout << "D"; }
S(const S&) { std::cout << "C"; }
S& operator=(const S&) { std::cout << "A"; return *this; }
};
void byValue(S) {}
void byRef(const S&) {}
int main() {
S a;
S b = a;
S c;
c = a;
byValue(a);
byRef(a);
}S a; → D. S b = a; builds a new object → copy constructor C. S c; → D. c = a; both exist → assignment A. byValue(a) copies the argument → C. byRef(a) makes no copy.
False. It copies member by member, so a pointer member's address is copied (a shallow copy): both objects share one buffer, which leads to a double-free. A deep copy needs your own copy constructor.
e->a equivalent to, written the long way?(*e).a — dereference the pointer e with *, then use the dot to access the member. -> just combines those two steps.
-> built-in or overloaded — and when does overloading come in?For a raw pointer it's built-in. Overloading operator-> comes in for your own classes — smart pointers (unique_ptr/shared_ptr), iterators, and custom wrapper types define it so they can be used like pointers.
operator-> return, and what does the compiler do with it?It returns a pointer. The compiler then applies the real -> to that returned pointer to reach the member — so w->print() becomes w.operator->()->print(). (If it returned another wrapper, -> keeps chaining until it hits a raw pointer.)
-> on a smart pointer even though it's an object, not a raw pointer?Because the smart pointer overloads operator-> (and operator*). So p->method() calls p.operator->() to get the raw pointer, then applies -> to that — letting an object behave like a pointer.
struct P { int x = 3; };
int main() {
P obj;
P* raw = &obj;
auto sp = std::make_unique<P>();
sp->x = 7;
raw->x += 1;
std::cout << obj.x << " " << (*sp).x;
}raw points at obj, so raw->x += 1 changes obj.x to 4. sp->x = 7 works because unique_ptr overloads ->; (*sp).x is the long way to read the same member.
operator-> must return a pointer, or an object that itself has operator->.True. The compiler applies -> again to whatever it returns, and keeps going until it reaches a raw pointer. That's how smart pointers and iterators support p->member.
size() and capacity()?size() is how many elements are currently stored. capacity() is the total number of slots currently allocated before the vector must reallocate. The spare room is capacity() - size(); always capacity() >= size().
push_back exceeds capacity, and why is it still amortized O(1)?It allocates a new, larger block (~1.5-2x), moves/copies all existing elements over, and frees the old block. Because capacity grows geometrically, reallocations get rarer as the vector grows: total copy work over N inserts sums to ~N (not N^2), so the average cost per push_back is constant — amortized O(1) — even though one call can be O(n).
reserve() optimize?It pre-allocates capacity up front (to your expected final size), so a run of push_backs doesn't keep triggering reallocations — each of which would copy/move all elements and destroy the old ones. One allocation instead of many.
push_back and emplace_back?push_back(obj) takes an already-built object and copies/moves it into the vector. emplace_back(args...) constructs the element directly inside the vector from the constructor arguments — no temporary object, so no extra copy/move + destroy.
v[i] and v.at(i)?v[i] does no bounds checking — an out-of-range index is undefined behaviour. v.at(i) checks the index and throws std::out_of_range if it's invalid (safer, slightly slower).
Because growth moves the elements to a new memory block and frees the old one. Any iterator/pointer/reference still pointing at the old block now points to freed memory (dangling) — using it is undefined behaviour.
std::vector<int> v = {10, 20, 30};
v.push_back(40);
v.emplace_back(50);
std::cout << v.size() << " " << v[4] << " " << v.front() + v.back();Three elements, then push_back(40) and emplace_back(50) make five. v[4] is the fifth element, 50, and front() + back() is 10 + 50 = 60.
int* p = &v[0];, then a push_back made the vector reallocate. What is p now?Dangling. On reallocation the vector moves its elements to a new, bigger block and frees the old one, so p still points into freed memory. Using it is undefined behaviour.
static on a local variable, and what stays the same?The lifetime changes: it now lives for the whole program and keeps its value between calls. The scope stays the same: only code inside that function can use the name. It behaves like a private global.
static Logger L; need a hidden flag but static int n = 0; does not?0 is a constant the compiler can store in the program file, so it is already there before main. A constructor is code, and the compiler can't run your program, so it runs at run time. The declaration line is reached on every call, so a hidden flag makes sure it only runs the first time.
int counter() { static int n = 0; return ++n; }
int main() {
for (int i = 0; i < 3; i++) std::cout << counter() << ' ';
}It prints 1 2 3. n is created once with value 0 and is not re-initialized on later calls, so each call increments the same variable.
arr safe with static int arr[3] but not with int arr[3]?A plain local array lives in the stack frame, which is destroyed when the function returns, so the returned pointer dangles. A static array lives in the static area until the program ends, so the pointer stays valid.
static Logger L; is never called. Is the constructor ever run? Is the destructor?Neither. The compiler decides from the initializer alone that a flag is needed. The check runs inside the function, so if the function never runs, the flag stays 0, the constructor never runs, and no destructor is registered.
static Logger L; at the same time. What is safe and what is not?Since C++11 the construction is thread-safe: one thread builds it and the other waits. But later changes to the object or to a static int (for example ++n) are not protected, so they are a data race. Use std::atomic or a mutex.
static Logger L; and static Logger* P = new Logger(); at program exit?L is destroyed at exit, so ~Logger() runs. With P only the pointer is static; the Logger object is on the heap, so its destructor never runs and the memory is not freed by your code.
int next() { static int n = 10; n += 5; return n; }
int main() {
std::cout << next() << " ";
std::cout << next() << " ";
std::cout << next();
}n is created once with 10 and keeps its value between calls, so each call adds 5 to the same variable: 15, 20, 25.
static local variable by name.False. static changes the variable's lifetime, not its scope. The name is still visible only inside its function.