在 C++ 中是否可以有一个“通用”模板参数，它可以是非类型模板参数或类型？
[英]Is it possible to have a "generic" template parameter in C++, that can be either a non-type template parameter or a type?
问题描述
In C++, it's possible to use a type as a template parameter, eg:
在 C++ 中，可以使用类型作为模板参数，例如：
template <typename T>
void MyFn();
It's also possible to use a non-type as a template parameter in some cases, eg:在某些情况下，也可以使用非类型作为模板参数，例如：
template <int64_t T>
void MyFn2();
My question is whether it's possible to have a "generic" template parameter that can be both?我的问题是是否有可能有一个“通用”模板参数可以是两者？ Like:喜欢：
template <TypenameOrint64_t T>
void MyFn3();
such that both MyFn3<42> and MyFn3<double> would be acceptable.这样MyFn3<42>和MyFn3<double>都是可以接受的。
An example of how I might use this:我如何使用它的一个例子：
template <typename ValType, ValType Head, ValType ...Tail>
struct ListS{
template <typename OutType, template <ValType ArgType> class Fn>
using MapHead = ListS<OutType, Fn<Head>::val, Tail...>;
};
template<int64_t N>
struct SquareS{
static constexpr const int64_t val = N * N;
};
using Sqrd = ListS<int64_t, 3, 4>::MapHead<int64_t, SquareS>;
static_assert(std::is_same<Sqrd, ListS<int64_t, 9, 4>>::value, "Values don't match");
The above is a very rough sketch of a compile-time list of values along with a single compile-time "function" on it.上面是一个非常粗略的编译时值列表以及一个编译时“函数”。 Would it be possible to make something like that also support lists of types, not just lists of non-type template param compatible values, without just duplicating all the code?是否有可能使类似的东西也支持类型列表，而不仅仅是非类型模板参数兼容值的列表，而不仅仅是复制所有代码？
2 个解决方案
解决方案1
3 已采纳 2020-02-01 13:21:03
Is it possible to have a “generic” template parameter in C++, that can be either a non-type template parameter or a type?
Short answer: no.简短的回答：没有。
Long answer.长答案。
No. The best I can imagine to mix types and values is wrap values in types, using std::integral_constant , by example.不。我能想象的最好的混合类型和值是将值包装在类型中，例如使用std::integral_constant 。
So, your desired code, could be written (C++17) almost as follows所以，你想要的代码，几乎可以写成（C++17）如下
#include <utility>
template <typename ...>
struct ListS;
template <typename ValType, ValType Head, ValType ...Tail>
struct ListS<std::integral_constant<ValType, Head>,
std::integral_constant<ValType, Tail>...>
{
template <template <auto> class Fn, typename OutType = ValType>
using MapHead = ListS<std::integral_constant<OutType, Fn<Head>::value>,
std::integral_constant<OutType, Tail>...>;
};
template <auto N>
struct SquareS : public std::integral_constant<decltype(N), N*N>
{ };
int main ()
{
using T1 = ListS<std::integral_constant<long long, 3ll>,
std::integral_constant<long long, 4ll>>;
using T2 = T1::MapHead<SquareS>;
using T3 = ListS<std::integral_constant<long long, 9ll>,
std::integral_constant<long long, 4ll>>;
static_assert( std::is_same_v<T2, T3> );
}
Pre C++17 you can't use auto for the type of the template values so you should make some simple corrections.在 C++17 之前，您不能将auto用于模板值的类型，因此您应该进行一些简单的更正。
解决方案2
2 2020-02-01 12:35:06
You could use function overloading and the auto type deduction that came with C++17 to accomplish something similar.您可以使用C++17附带的函数重载和auto类型推导来完成类似的工作。
template<typename myType>
auto myFn3(myType value){
return value;
}
template<auto value> //takes any non-type parameter
auto myFn3(){
return value;
}
int main(){
auto test1_normal = myFn3(3);
auto test1_cast = myFn3<double>(3); //able to perform a cast
auto test1_auto = myFn3<3>();
return 0;
}
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