移动在 return 上调用的构造函数而不是复制

Question

今天我发现这段代码不能像我期望的那样工作。根据我对 L 值的知识，应该调用复制构造函数，而对于 R 值，它应该选择移动构造函数。否则 std::move 的目的是什么，它实际上只转换为 R 值。我原以为 return obj 会调用复制构造函数，但它调用了 move。我知道复制在这里没有用，但这是关于规则的。如果我的复制构造函数有副作用怎么办，这就是我的情况（我知道它不应该，但从技术上讲它可以 - 例如这里：std::cout 调用）。标准中是否有任何允许此类行为的内容？还有我怎样才能强制复制？

#include <iostream>

class X
{
public:
    X() = default;

    X(const X& r): i(r.i)
    {
        std::cout << "copy ctor" << std::endl;
    }

    X(const X&& r): i(r.i)
    {
        std::cout << "move ctor" << std::endl;
    }
    int i = 0;
};

X foo()
{
    X obj;
    obj.i = 10;
    return obj;
}

int main()
{
    X x = foo();
}

move ctor

move ctor

Answer 1

来自 cppreference（强调我的）：

If [the returned expression] is an lvalue expression and the conditions for copy elision are met, or would be met, except that [the returned expression] names a function parameter, then overload resolution to select the constructor to use for initialization of the returned value is performed twice: first as if [the returned expression] were an rvalue expression (thus it may select the move constructor or a copy constructor taking reference to const), and if no suitable conversion is available, overload resolution is performed the second time, with lvalue [returned expression] (so it may select the copy constructor taking a reference to non-const). The above rule applies even if the function return type is different from the type of [the returned expression] (copy elision requires same type)

长话短说，return 会在有意义的时候暗中尝试移动您 return 的内容。它只会作为最后的手段进行复制（例如，没有可用的移动构造函数）。

移动在 return 上调用的构造函数而不是复制

move constructor called on return instead of copy

move-semantics

c++11