初始化具有函数返回值的对象时未复制的复制构造函数

Question

Consider the following code:

#include <iostream>

using namespace std;

class A
{
    public:
        int a;
        A(): a(5)
        {
           cout << "Constructor\n";
        }
        A(const A &b)
        {
            a = b.a;
            cout << "Copy Constructor\n";
        }
        A fun(A a)
        {
            return a;
        }
};

int main()
{
    A a, c;
    A b = a.fun(c);
    return 0;
}

The output of the above code with g++ file.cpp is:

Constructor
Constructor
Copy Constructor
Copy Constructor

The output of the above code with g++ -fno-elide-constructors file.cpp is:

Constructor
Constructor
Copy Constructor
Copy Constructor
Copy Constructor

I know Return Value Optimization. My question is which call to copy constructor is elided(temporary object during returning or returned object being copied to b)?

If the elided copy constructor is the one used for creating b, then how is b created at all (because there is no constructor call in this case also)?

If I replace the line A b = a.fun(c); with a.fun(c) and compile using the first method or even the second method, then also the copy constructor is being called 2 times . So, if in the case explained in the previous paragraph, the temporary object's copy constructor is elided, then why isn't it elided in this case?

Answer 1

#include <iostream>

using namespace std;

class A
{
public:
    int a;
    A(): a(5)
    {
        cout << "Constructing: " << (void *)this << std::endl;
    }
    A(const A &b)
    {
        a = b.a;
        cout << "Copy Constructor: " << (void *)this << " from " << (void *)&b << std::endl;
    }
    A fun(A a)
    {
        return a;
    }
};

int main()
{

    A a, c;
    A b = a.fun(c);

    std::cout << "a:" << (void *)&a << std::endl <<
              "b:" << (void *)&b << std::endl <<
              "c:" << (void *)&c << std::endl;
    return 0;
}

Yields:

Constructing: 0x7fffbb377220
Constructing: 0x7fffbb377210
Copy Constructor: 0x7fffbb377230 from 0x7fffbb377210
Copy Constructor: 0x7fffbb377200 from 0x7fffbb377230
a:0x7fffbb377220
b:0x7fffbb377200
c:0x7fffbb377210

So it constructs a , constructs c , copies c to an intermediate (argument a of the function), and then copies the intermediate directly into b , skipping the typical copying of a to a return intermediate. This is even better demonstrated if you pass by value (change to A fun(const A& a) :

Constructing: 0x7fff8e9642b0
Constructing: 0x7fff8e9642a0
Copy Constructor: 0x7fff8e964290 from 0x7fff8e9642a0
a:0x7fff8e9642b0
b:0x7fff8e964290
c:0x7fff8e9642a0

a is constructed, c is constructed, c is copied directly to b, despite b not being passed to fun!

Answer 2

The copy that is elided is the copy of the temporary return value into b . Without elision the return value is initialized from a and copied to b . Instead, the temporary that would otherwise hold the return value is constructed into b and initialized with a . [class.copy]/31:

when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move

You can observe this if you add an additional output in fun :

A fun(A a)
{
    cout << "fun!" << endl;
    return a;
}

Then with the elision you'll get

[…]
fun!
Copy Constructor

And without:

[…]
fun!
Copy Constructor
Copy Constructor