C++ 复制构造函数和默认构造函数

Question

i am learning c++ and have found a output which i dont really understand.

#include <iostream>
using namespace std;

class A{
    public:
        A(){ cout << "A+" << endl;}
        A(const A&){ cout << "A(A)" << endl;}
        ~A(){cout << "A-" << endl;}
};

class B{
    public:
        B(){ cout << "B+" << endl;}
        B(const B&){cout << "B(B)" << endl;}
        ~B(){cout << "B-" << endl;}
    private:
        A a;
};

class C : public A{
    public:
        C(const B& b) : b1(b) { cout << "C+" << endl;}
        ~C(){ cout << "C-" << endl;}
    private:
        B b1,b2;
};

void test(A a){
    A m(a);
}

int main(){
    B b;
    C c(b);
    test(c);
    return 0;

}

the output is

A+
B+
A+
A+
B(B)
A+
B+
C+
A(A)
A(A)
A-
A-
C-
B-
A-
B-
A-
A-
B-
A-

I mean the first one, B goes to default sees a we got a member from type A and goes to A thats the A+ than goes back to B and print B+ . Thats it with B b; than C c(b) it goes to C, see its public A goes to A and print A+ than goes back see we got a Member B b1,b2 goes to B and B have a member A and goes agean to A and print A+ and than i dont understand why B(B) ? after this B(B)i dont understand anything.. i try it to debugg but it didnt help me very much, maybe someone can explain why this works like this?

Answer 1

If I have understood correctly your question you are trying to understand the output

A+
A+
B(B)
A+
B+
C+

that corresponds to this declaration

C c(b);

The class C has base class A

class C : public A{

So the constructor of the class A is called

A+

then the data member b1 is created

C(const B& b) : b1(b) { cout << "C+" << endl;}

The class B in turn has data member A

class B{
    public:
        B(){ cout << "B+" << endl;}
        B(const B&){cout << "B(B)" << endl;}
        ~B(){cout << "B-" << endl;}
    private:
        A a;
};

So when the copy constructor of the class B is called the data member a is created

A+
B(B)

The class C has one more data member of the class B. It is the data member b2. So these constructors are called

A+
B+

And at last the body of the constructor C gets control

C+

Destructors get the control in the reverse order relative to the order of creating objects.

So the destructors output of the object c looks the following way

C-
B-
A-
B-
A-
A-

You can make the program output more clear with minor changes of the program.

For example

#include <iostream>
using namespace std;

class A{
    public:
        A(){ cout << "A+" << endl;}
        A(const A&){ cout << "A(A)" << endl;}
        ~A(){cout << "A-" << endl;}
};

class B{
    public:
        B() : i( n++ ) { cout << "B+" << ' ' << i << endl;}
        B(const B&) : i( n++ ) {cout << "B(B)" << ' ' << i << endl;}
        ~B(){cout << "B-" << ' ' << i << endl;}
    private:
        size_t i;
        static size_t n;
        A a;
};

size_t B::n;

class C : public A{
    public:
        C(const B& b) : b1(b) { cout << "C+" << endl;}
        ~C(){ cout << "C-" << endl;}
    private:
        B b1,b2;
};

void test(A a){
    A m(a);
}

int main(){
    B b;

    std::cout << '\n';

    C c(b);

    std::cout << '\n';

    test(c);

    std::cout << '\n';
}

The program output of this updated program is

A+
B+ 0

A+
A+
B(B) 1
A+
B+ 2
C+

A(A)
A(A)
A-
A-

C-
B- 2
A-
B- 1
A-
A-
B- 0
A-

Answer 2

Lets take a closer look at the C constructor (slightly reformatted):

C(const B& b)
    :
    b1(b)
{
    cout << "C+" << endl;
}

First the A constructor will be invoked, as it's a base-class for C . That will print A+ .

Then the b1 member will be copy-constructed, which will print first A+ because of the B::a member, followed by B(B) in the B copy-constructor body.

Then the b2 member will be default constructed, which will print A+ (again because of the B::a member) followed by B+ .

Then the C constructor body will run which will print C+ .

---

The C constructor is really equivalent to this (with comments added):

C(const B& b)
    : A(),      // Prints A+
      b1(b),    // Prints A+ and B(B)
      b2()      // Prints A+ and B+
{
    cout << "C+" << endl;    // Prints C+
}

Hopefully this will make it easier to see what's going on.