C++ 检测类型是否有模板参数 [英] C++ detect if type has template parameter

问题描述

我想统一一个接口来处理模板化和非模板化类型.有没有办法确定类型(例如类或函数指针)是否依赖于模板参数?

I would like to unify an interface to work with both templated and non-templated types. Is there a way to determine whether a type, such as a class or a function pointer is depends on a template parameter?

例如:

struct foo {};
template<typename T>  struct bar {};

// This works if the template parameter is provided
template<typename> struct is_templated : false_type {};
template<template<typename...> class Obj, typename...Args>
struct is_templated<Obj<Args...>> : true_type {};
template<typename T> constexpr auto is_templated_v = is_templated<T>::value;

在这种情况下，is_template_v<foo> 是假的，而 is_template_v<bar<int>> 是真的，但我不能用 is_template_v.或者，如果我定义

In this case, is_template_v<foo> is false and is_template_v<bar<int>> is true, but I can't just deduce anything with is_template_v<bar>. Alternatively, if I define

template<template<typename...> class> 
struct temp_check : true_type {};

然后 temp_check 是完全有效的，但我不知道如何类似地检查 foo.如果它是有效的 C++，则需要这样的东西

Then temp_check<bar> is perfectly valid, but I don't know how I would analogously check foo. What is needed is something like this if it were valid C++

template<template<> class A> struct temp_check<A> : false_type {};

是否有某种机制可以同时检查两者?

Is there some mechanism that could check both?

推荐答案

我会使用重载集的力量:

I'd use the power of overload sets:

#include <iostream>
#include <type_traits>

struct foo {};
template<typename T>  struct bar {};

template<template<class ...> class T, class... TArgs>
constexpr bool is_template() { return true; }

template<class T>
constexpr bool is_template() { return false; }

int main()
{
    std::cout << is_template<foo>() << '\n'; // 0
    std::cout << is_template<bar>() << '\n'; // 1
}

让给用户:使用模板函数提供一个特征;)

Let to the user: use the template function to provide a trait ;)

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