c++ - 使用移动时不会发生复制省略

Question

为什么这个例子打印:

#include <iostream>

struct X
{
    X() = default;
    X(X const&) { std::cout << "copy-constructor\n"; }
    X(X&&) { std::cout << "move-constructor\n"; }

    X& operator=(X)
    {
        return *this;
    }
};

int main() 
{
    X x, y;
    std::cout << "assign from prvalue calls the ";
    x = X{};
    std::cout << "\nassign from xvalue calls the ";
    x = std::move(y);
}

assign from prvalue calls the
assign from xvalue calls the move-constructor

X{} 和 std::move(y) 都是右值，那么为什么只分配给 X{} 会导致复制省略？

Answer 1

复制省略在第一种情况下有效，因为您正在从一个临时值初始化赋值运算符的参数； temporary 可以省略，直接构造参数。用标准的话来说，省略的标准之一是:

when a temporary class object that has not been bound to a reference would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move

在第二种情况下，您不是从临时对象而是从现有对象初始化它。它已经在与目标不同的位置构建，因此无法进行上述优化。