检测类型是否为 std::tuple?
目前我有两个功能:
template<typename Type> bool f(Type* x);
template<typename... List> bool f(std::tuple<List...>* x);
有什么方法可以将这两个函数与一个额外的模板参数合并，该参数指示传递的类型是否为元组?
Is there any way to merge these two functions with an extra template parameter that indicates whether the passed type is a tuple ?
template<typename Type, bool IsTuple = /* SOMETHING */> bool f(Type* x);
推荐答案
当然，使用 is_specialization_of(链接取自 这里):
template<typename Type, bool IsTuple = is_specialization_of<Type, std::tuple>::value>
bool f(Type* x);
然而，问题是，你真的想要这样吗?通常，如果您需要知道一个类型是否是元组，您需要对元组进行特殊处理，这通常与它的模板参数有关.因此，您可能希望坚持使用重载版本.
The question is, however, do you really want that? Normally, if you need to know if a type is a tuple, you need special handling for tuples, and that usually has to do with its template arguments. As such, you might want to stick to your overloaded version.
既然你提到你只需要一小部分专门化，我建议重载，但只适用于小的特殊部分:
Since you mentioned you only need a small portion specialized, I recommend overloading but only for the small special part:
template<class T>
bool f(T* x){
// common parts...
f_special_part(x);
// common parts...
}
与
template<class T>
void f_special_part(T* x){ /* general case */ }
template<class... Args>
void f_special_part(std::tuple<Args...>* x){ /* special tuple case */ }
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