C ++默认，复制和升级构造函数

Question

I have the following code [This is an interview question]:

#include <iostream>
#include <vector>

using namespace std;

class A{
public:
    A(){
        cout << endl << "base default";
    }
    A(const A& a){
        cout << endl << "base copy ctor";
    }
    A(int) { 
        cout << endl << "base promotion ctor";
    }
};

class B : public A{
public:
    B(){
         cout << endl << "derived default";
    }
    B(const B& b){
         cout << endl << "derived copy ctor";
    }
    B(int) {
         cout << endl << "derived promotion ctor";
    }
};

int main(){

    vector<A> cont;
    cont.push_back(A(1));
    cont.push_back(B(1));
    cont.push_back(A(2));

        return 0;
    }

The output is :

base promotion ctor
base copy ctor
base default
derived promotion ctor
base copy ctor
base copy ctor
base promotion ctor
base copy ctor
base copy ctor
base copy ctor

I am having trouble understanding this output, specifically why base default is called once and the last 3 copy ctor. Can someone please explain this output ?

Thanks.

Answer 1

The base default constructor is called once from the line

cont.push_back(B(1));  

All your B constructors call the default A constructor. The last two copy constructors are because of a vector re-allocation. For example, if you added

cont.reserve(3);

before the push_back s, they'd go away.

The one before that is a copy of the temporary A(2) in your final push_back .