constructor or copy constructor?

Question

In the book Generic Programming and the STL (Chinese edition), it says:

X x = X() will call the copy constructor.

It seems a little weird to me. And I write a test program like this

#include <iostream>

class Test {

public:

    Test() {
        std::cout << "This is ctor\n";
    }

    Test(const Test&) {
        std::cout << "This is copy-ctor\n";
    }

};

int main(int argc, char** argv)
{

    Test t = Test();
    return 0;
}

The output is "This is ctor". ok, now I'm confused, which is right?

Answer 1

Nomimally yes, a temporary is default-constructed, and then the copy constructor is invoked to copy it into your object t.

However, in practice the copy can be optimised out — even though it has side effects (the console output):

[n3290: 8.5/16]: [..] In certain cases, an implementation is permitted to eliminate the copying inherent in this direct-initialization by constructing the intermediate result directly into the object being initialized; see 12.2, 12.8.

And (in conjunction with the example given in the same clause):

[n3290: 12.2/2]: [..] An implementation might use a temporary in which to construct X(2) before passing it to f() using X’s copy constructor; alternatively, X(2) might be constructed in the space used to hold the argument. [..]

But the copy constructor does still have to exist, even though it might not be invoked.

Anyway, if you compile with optimisations turned off (or, with GCC, possibly -fno-elide-constructors), you will see:

This is ctor
This is copy-ctor

Answer 2

In theory, X x = X() will call the default constructor to create a temporary object, and copy that into x using the copy constructor.

In practice, compilers are allowed to skip the copy construction part and default-construct x directly (which, as David points out in his comment, still requires the copy constructor to be syntactically accessible, though). Most compilers do that at least when optimizations are enabled.