Why call std::move on a variable that is already T&&?

A StackOverflow question asks why it is necessary to call std::move on a parameter when implementing a move constructor, given that the parameter is already declared T&&?

Named rvalue references are lvalues. Unnamed rvalue references are rvalues.

In other words, if you don’t explicitly move a into b X b = std::move(a); then you haven’t declared that you’ve finished with a, which is a named variable – so the compiler will call X’s copy constructor instead of X’s move constructor.

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Filed under C++, Programming

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