元编程:函数定义失败定义了一个单独的函数

2021-12-13 00:00:00 templates c++ template-meta-programming sfinae result-of

在 这个答案中，我根据类型的 is_arithmetic 属性定义了一个模板:

In this answer I define a template based on the type's is_arithmetic property:

template<typename T> enable_if_t<is_arithmetic<T>::value, string> stringify(T t){
    return to_string(t);
}
template<typename T> enable_if_t<!is_arithmetic<T>::value, string> stringify(T t){
    return static_cast<ostringstream&>(ostringstream() << t).str();
}

dyp 建议而不是 类型的 is_arithmetic 属性，即是否为类型定义了 to_string 作为模板选择标准.这显然是可取的，但我不知道怎么说:

dyp suggests that rather than the is_arithmetic property of the type, that whether to_string is defined for the type be the template selection criteria. This is clearly desirable, but I don't know a way to say:

如果 std::to_string 未定义，则使用 ostringstream 重载.

If std::to_string is not defined then use the ostringstream overload.

声明 to_string 条件很简单:

template<typename T> decltype(to_string(T{})) stringify(T t){
    return to_string(t);
}

这与我不知道如何构建的标准相反.这显然不起作用，但希望它传达了我正在尝试构建的内容:

It's the opposite of that criteria that I can't figure out how to construct. This obviously doesn't work, but hopefully it conveys what I'm trying to construct:

template<typename T> enable_if_t<!decltype(to_string(T{})::value, string> (T t){
    return static_cast<ostringstream&>(ostringstream() << t).str();
}

推荐答案

在上周的委员会会议上新鲜投票进入图书馆基础 TS:

Freshly voted into the library fundamentals TS at last week's committee meeting:

template<class T>
using to_string_t = decltype(std::to_string(std::declval<T>()));

template<class T>
using has_to_string = std::experimental::is_detected<to_string_t, T>;

然后将 has_to_string 上的 dispatch 和/或 SFINAE 标记为您的心意.

Then tag dispatch and/or SFINAE on has_to_string to your heart's content.

您可以参考TS的当前工作草案is_detected 和朋友是如何实现的.它与@Yakk 的回答中的 can_apply 非常相似.

You can consult the current working draft of the TS on how is_detected and friends can be implemented. It's rather similar to can_apply in @Yakk's answer.

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