Question

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Why copy constructor is not called in this case?

In the following code, I constructed three variables, a1, a2 and a3.

There’s a example in C++ Primer p.476:

string empty_copy = string();//copy-initialization

Is there anyone can help me explain

1)why a1 and a2 are not constructed by copy constructor and

2)what’s the difference between initialization a2 in my code and empty_copy in the book?

Thanks so much!

#include<iostream>
using namespace std;
class A{
public:
    A(){}
    A(int v){}
    A(const A&x){
        cout<<"copy constructor"<<endl;
    }
};
A generateA(){
    return A(0);
}
int main(){
        cout<<"First:"<<endl;
        A a1=generateA();

        cout<<"Second:"<<endl;
        A a2=A(0);

        cout<<"Third:"<<endl;
        A a3=a1;
    return 0;
}

The out put is (under Visual Studio 2010 in Win7 and g++ in Ubuntu10.10):

First:
Second:
Third:
copy constructor

Answer 1

This is copy elision due to Return value optimization.
Compilers are allowed to optimze away generation of copies by applying such optimizations.

A a1=generateA(); 
A a2=A(0); 

In the above both cases the compiler can eliminate the creation of a temporary object which is created to hold the return value.

A a3=a1;

Involves a already existing named object a1 which is used for construction of a3 this involves a copy constructor call which the compiler must make and cannot optimize.

EDIT: To answer the Q in comment.

You can tell the compiler to not apply this optimization by using the following options during compilation:

For GCC:

-fno-elide-constructors

For MVSC:

/Od