c++ - R值的C ++隐式转换(C++ Implicit Conversion from R-Value)

Question

I'm trying to define an implicit conversion operator that will move a wrapper type into the implicit type as such:

(我正在尝试定义一个隐式转换运算符，该运算符会将包装器类型移入隐式类型，例如：)

struct SomeStruct
{
    int a;
    int b;
    int c;
};

struct AnyStruct
{
    template <typename T>
    operator T&() &
    {
        return Get<T>();
    }

    template <typename T>
    operator T() &&
    {
        return std::move(Get<T>());
    }

    // details
    // ...
};

int main(void)
{
    SomeStruct s1 { 1, 2, 3 };
    AnyStruct s2 = std::move(s1);

    SomeStruct& s3 = s2;           // implicit conversion working
    SomeStruct s4 = std::move(s2); // implicit conversion not working (copying)
}

The first implicit conversion works and calls the right cast operator operator T&() & .

(第一个隐式转换起作用并调用正确的强制转换运算operator T&() & 。)

I would have expected the second implicit conversion to call the right cast operator operator T() && and move s2 into s4 , but it calls the cast operator operator T&() & instead and actually copy the struct.

(我本来希望第二次隐式转换调用正确的强制转换运算operator T() &&并将s2移至s4 ，但它会调用s2转换运算operator T&() &并实际上复制该结构。)

What am I doing wrong?

(我究竟做错了什么？)

  ask by sturcotte06 translate from so

Answer 1

等待大神答复