Check if a type is passed in variadic template parameter pack

南笙酒味 提交于 2019-11-30 13:50:10

You're looking for std::disjunction. It's specified in N4564 [meta.logical].

#include <type_traits>

template<typename T, typename... Ts>
constexpr bool contains()
{ return std::disjunction_v<std::is_same<T, Ts>...>; }

static_assert(    contains<int,      bool, char, int, long>());
static_assert(    contains<bool,     bool, char, int, long>());
static_assert(    contains<long,     bool, char, int, long>());
static_assert(not contains<unsigned, bool, char, int, long>());

Live demo


Or, adapted to a struct

template<typename T, typename... Ts>
struct contains : std::disjunction<std::is_same<T, Ts>...>
{};

Or, using fold expressions

template<typename T, typename... Ts>
struct contains : std::bool_constant<(std::is_same<T, Ts>{} || ...)>
{};

Live demo

If you are bound to C++11 you cannot use std::disjunction nor fold-expressions. However, it is rather straightforward to roll your own any_is_same:

template<typename same, typename first,typename...more> 
struct any_is_same {
    static const bool value = std::is_same<same,first>::value || 
                              any_is_same<first,more...>::value; 
};

template<typename same,typename first> 
struct any_is_same<same,first> : std::is_same<same,first> {};


int main(){
    std::cout << any_is_same<int, int,double,float>::value << "\n";
    std::cout << any_is_same<std::string, int,double,float>::value << "\n";
}

Live Demo

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