Can someone explain the output of this c++ program?

Question

Why is the output foo3 equal to 3 ? i would suggest, when bar(foo1) is called, the function bar creates a copy of foo1 on the stack, so it's value is equal to 0, when this value is returned, the copy-constructor for foo3 increments the value again so it should be 2?

Thanks in advance.

this is my code:

#include <iostream>
struct Foo {
    Foo()
        : x(0)
    {
    }
    Foo(const Foo& foo)
        : x(foo.x + 1)
    {
    }
    int x;
};

Foo bar(Foo foo)
{
    foo.x++;
    return foo;
}

int main()
{
    Foo foo1;
    Foo foo2 = foo1;
    std::cout << "A:" << foo1.x << std::endl;
    std::cout << "B:" << foo2.x << std::endl;
    Foo foo3 = bar(foo1);
    std::cout << "C:" << foo3.x << std::endl;
}

output:

A:0
B:1
C:3

Answer 1

I believe there are three copy constructors at work here, the line foo2 = foo1, the passing of foo1 into bar, and the returning of foo1 from bar.

Modifying your code makes it clear what is happening:

#include <iostream>
struct Foo {
    Foo()
        : x(0)
    {
        std::cout << "Constructor called" << std::endl;
    }
    Foo(const Foo& foo)
        : x(foo.x + 1)
    {
        std::cout << "Copy constructor called" << std::endl;
    }
    int x;
};

Foo bar(Foo foo)
{
    std::cout << "B2:" << foo.x << std::endl;
    foo.x++;
    return foo;
}

int main()
{
    Foo foo1;
    Foo foo2 = foo1;
    std::cout << "A:" << foo1.x << std::endl;
    std::cout << "B:" << foo2.x << std::endl;
    Foo foo3 = bar(foo1);
    std::cout << "C:" << foo3.x << std::endl;
}

Output:

Constructor called
Copy constructor called
A:0
B:1
Copy constructor called
B2:1
Copy constructor called
C:3