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C ++ SFINAE void_t无法正常工作

[英]C++ SFINAE void_t not working

I tried running following code, which should rely on void_t trick, where more specialized class template should be chosen (in this case second one) 我尝试运行以下代码,该代码应依赖void_t技巧,应在其中选择更专业的类模板(在本例中为第二个模板)

#include <iostream>
#include <type_traits>

template <class ...>
using void_t = void;

template <class T, class = void>
struct is_incrementable : public std::false_type { };
template <class T>
struct is_incrementable<T, void_t<decltype(++(std::declval<T>()))>> : public std::true_type { };

int main()
{
    std::cout << std::boolalpha;
    std::cout << is_incrementable<int>::value << std::endl;
    return 0;
}

I am using MSVC 2015. However, the result of is_incrementable<int>::value is false. 我正在使用is_incrementable<int>::value 。但是, is_incrementable<int>::value为false。 Is there anything wrong with my code or is there problem with my compiler? 我的代码有什么问题还是我的编译器有问题?

std::declval<T> has return type T&& , so std::declval<int>() is an rvalue of type int . std::declval<T>具有返回类型T&& ,因此std::declval<int>()int类型的右值 The result of "false" is telling you that an int rvalue is not incrementable, which is correct. 结果“ false”告诉您int rvalue不可递增,这是正确的。

You can replace std::declval<T> with std::declval<T&> to get the program to tell you whether an lvalue of type T is incrementable. 您可以将std::declval<T>替换为std::declval<T&>以使程序知道T类型的左值是否可递增。 If you make this change to your program, it should print "true". 如果对程序进行了更改,则应打印“ true”。

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