Why does calling std::move on a const object call the copy constructor when passed to another object?

Question

Why does calling std::move on a const object call the copy constructor when passed to another object? Specifically, the code

Answer 1

The type of std::move(x) is Foo const&& which can't bind to Foo&&. The reasoning is the same as for a T const& not being able to bind to a T&. You can, however, have a constructor taking a Foo const&&. Most likely you won't be able to really move the corresponding object's data but, e.g., in your example there no data, i.e., the following code works OK:

#include <iostream>

struct Foo {
    Foo() = default;
    Foo(Foo &&) { std::cout << "Move\n"; }
    Foo(Foo const&&) { std::cout << "Move const\n"; }
    Foo(Foo const &) = delete;
};

int main() {
    Foo const x; Foo y(std::move(x)); 
}

Answer 2

The type of the result of calling std::move with a T const argument is T const&&, which cannot bind to a T&& parameter. The next best match is your copy constructor, which is deleted, hence the error.

Explicitly deleteing a function doesn't mean it is not available for overload resolution, but that if it is indeed the most viable candidate selected by overload resolution, then it's a compiler error.

The result makes sense because a move construction is an operation that steals resources from the source object, thus mutating it, so you shouldn't be able to do that to a const object simply by calling std::move.