使用decltype / SFINAE检测操作员支持 [英] Detect operator support with decltype/SFINAE

问题描述

（稍微）过时的文章探讨了使用 decltype 以及SFINAE来检测类型是否支持某些运算符，例如 == 或< 。

A (somewhat) outdated article explores ways to use decltype along with SFINAE to detect if a type supports certain operators, such as == or <.

以下是检测类是否支持< 运算符的示例代码：

Here's example code to detect if a class supports the < operator:

template <class T>
struct supports_less_than
{
    static auto less_than_test(const T* t) -> decltype(*t < *t, char(0))
    { }

    static std::array<char, 2> less_than_test(...) { }

    static const bool value = (sizeof(less_than_test((T*)0)) == 1);
};

int main()
{
    std::cout << std::boolalpha << supports_less_than<std::string>::value << endl;
}

这会输出 true ，因为当然 std :: string 支持< 运算符。但是，如果我尝试使用不支持< 运算符的类，则会出现编译错误：

This outputs true, since of course std::string supports the < operator. However, if I try to use it with a class that doesn't support the < operator, I get a compiler error:

error: no match for ‘operator<’ in ‘* t < * t’

所以SFINAE不工作。我试过这个在GCC 4.4和GCC 4.6，和两个展示相同的行为。那么，是否可以使用SFINAE以这种方式检测类型是否支持某些表达式？

So SFINAE is not working here. I tried this on GCC 4.4 and GCC 4.6, and both exhibited the same behavior. So, is it possible to use SFINAE in this manner to detect whether a type supports certain expressions?

推荐答案

您需要将less_than_test函数作为模板，因为SFINAE表示替换失败不是错误，可能会在您的代码中失败选择。

You need to make your less_than_test function a template, since SFINAE stands for Substitution Failure Is Not An Error and there's no template function that can fail selection in your code.

template <class T>
struct supports_less_than
{
    template <class U>
    static auto less_than_test(const U* u) -> decltype(*u < *u, char(0))
    { }

    static std::array<char, 2> less_than_test(...) { }

    static const bool value = (sizeof(less_than_test((T*)0)) == 1);
};

int main()
{
    std::cout << std::boolalpha << supports_less_than<std::string>::value << endl;
}

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