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

#include <iostream>  struct Foo {     Foo() = default;     Foo(Foo && x) { std::cout << "Move" << std::endl; }     Foo(Foo const & x) = delete; };  int main() {     Foo const x; Foo y(std::move(x));  } 

fails to compile with the message:

g++ -std=c++14 test07.cpp -o test07 test07.cpp: In function 'int main()': test07.cpp:10:36: error: use of deleted function 'Foo::Foo(const Foo&)'      Foo const x; Foo y(std::move(x));                                      ^ test07.cpp:6:5: note: declared here      Foo(Foo const & x) = delete;      ^ Makefile:2: recipe for target 'all' failed make: *** [all] Error 1 

Certainly, I expect it to fail because we can't move a const value. At the same time, I don't understand the route that the code takes before it tries to call the copy constructor. Meaning, I know that std::move converts the element to an x-value, but I don't know how things proceed after that with respect to const.

Answer 1

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.