Why are constructors called if they are not inherited?

Question

code prints all constructors. I read that constructors are not inherited when we derive a class from another class. then why creation ccalls constructors from banda

class A
{
public:
  A()  { cout << "A constructor called" << endl; }
};

class B
{
public:
  B()  { cout << "B constructor called" << endl; }
};

class C: public B, public A  // Note the order
{
public:
  C()  { cout << "C constructor called" << endl; }
};

int main()
{
    C c;
    return 0;
}

Answer 1

When the document you are reading is not inherited, the constructors mean that if the class Adefines a constructor A::A(int x), then the child class Bwill not automatically have a constructor that accepts int.

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class A
{
public:
  A(int x)  { cout << "A constructor called" << endl; }
};

class C: public A
{
public:
  C()  
  : A(7)  /* compilation will fail without this line */
  { cout << "C constructor called" << endl; }
};

Answer 2

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struct A
{
   A(int) {}
};

struct B : A
{
   B() : A(0) {}
};

, :

B b(10);

A(int) B.

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  B() : A(0) {}
  

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  B() {}
  

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  B() : A() {}
  

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C, :

C()  { cout << "C constructor called" << endl; }

C() : B(), A() { cout << "C constructor called" << endl; }

B::B() A::A() C.

Answer 3

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Answer 4

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Answer 5

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Answer 6

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class A
{
    public: 
        A(int x) {}
};

class B: public A
{
};

int main(void)
{
    B b(5);
    return 0;
}

, A(int) . B, A(int)

class B: public A
{
     using A::A;
};

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Answer 7

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class A {
    public:
        A() {
            std::cout << "Constructor A" << '\n';
        }
};

class B : public A {
    public:
        B() {
            std::cout << "Constructor B" << '\n';
        }
};

class C : public B {
    public:
        C() {
            std::cout << "Constructor C" << '\n';
        }
};

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