类型不推论为r值引用：为什么不呢？ [英] Type not deduced to be r-value reference: why not?

问题描述

请考虑以下代码：

class CMyClass {};

template<class T>
void func(T&& param) {
        if (std::is_same<CMyClass, std::decay<T>::type>::value)
            std::cout << "param is a CMyClass\n";
        if (std::is_same<T, CMyClass&>::value)
            std::cout << "param is a CMyClass reference\n";
        else if (std::is_same<T, CMyClass&&>::value)
            std::cout << "param is a CMyClass r-value reference\n";
        else if (std::is_same<T, const CMyClass&>::value)
            std::cout << "param is a const CMyClass reference\n";
        else if (std::is_same<T, const CMyClass&&>::value)
            std::cout << "param is a const CMyClass r-value reference\n";
        else if (std::is_same<T, const CMyClass>::value)
            std::cout << "param is a constant CMyClass\n";
        else if (std::is_same<T, CMyClass>::value)
            std::cout << "param is a CMyClass\n";
        else
            std::cout << "param is not a CMyClass\n";
}


CMyClass mc3;
func(std::move(mc3));

此小程序的输出为

param is a CMyClass
param is a CMyClass

为什么不能将mc3的类型推论为r值引用？

Why has the type of mc3 not been deduced to be an r-value reference please?

推荐答案

我找不到

以下项的扣除规则：

template <class T>
void foo(T&& )

在通话中 foo（expr）是：

• 如果 expr 是类型 U 的左值，然后 T 推导为 U& 和类型 T& 是 U& ，这是因为引用崩溃了。


• 如果 expr 是类型为 U 的右值，则 T 推导为 U ，类型 T& 为 U& ，由于引用崩溃。



• If expr is an lvalue of type U, then T is deduced as U& and the type T&& is U&, due to reference collapsing.
• If expr is an rvalue of type U, then T is deduced as U the type T&& is U&&, due to reference collapsing.

在您的示例中， std :: move（mc3）是 CMyClass 类型的右值（特别是xvalue）。因此，将 T 推导为 CMyClass 。此检查：

In your example, std::move(mc3) is an rvalue (specifically an xvalue) of type CMyClass. Hence, T is deduced as CMyClass. This check:

else if (std::is_same<T, CMyClass&&>::value)
    std::cout << "param is a CMyClass r-value reference\n";

几乎不会像 T 那样真实。从不推论为右值引用类型。可以这样具体提供：

will almost never be true as T will never deduce as an rvalue reference type. It could be specifically provided as such:

func<CMyClass&&>(std::move(mc3));

但这是不太可能的用法。相反，您可以做的是检查：

but that's an unlikely usage. What you can do instead is check:

else if (std::is_same<T&&, CMyClass&&>::value)
//                    ~~~~

该参数是一个右值。确实，如果您只需要经常检查 T& ，就可以正确处理所有情况。

That will handle all cases where the argument is an rvalue. Indeed, if you simply always check for T&&, that will handle all of your cases properly.

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