C++ void_t

考虑如下代码:

#include <type_traits>
#include <iostream>
#include <vector>
using namespace std;

template <typename T, typename = void>
struct RangeConcept : std::false_type {};

template <typename T>
struct RangeConcept<T, 
                       std::void_t<
                                   decltype(  std::declval<T >().begin() ),
                                   decltype(  std::declval<T >().end() )
                                  >
                      >
    : std::true_type {};

template<typename Cont>
void show( Cont c){
    static_assert( RangeConcept< Cont >::value, "must be a range. means.. it has begin() and end()" );
}
int main() {
std::vector
<int> vec{1,2,3,4,5}; show(vec);
}

这就是C++17带来的void_t神器。写个静态断言无比容易。手写个Concept检查也不再是C++库开发者的专利了。

配合SNINAE和enable_if ,如下:

template<typename T>
void show(T& t, typename std::enable_if< RangeConcept<T>::value >::type* = 0){
    std::cout << "I am a Range" << std::endl;
}

template<typename T>
void show(T& t, typename std::enable_if< std::is_integral<T>::value >::type* = 0){
    std::cout << "I am a others" << std::endl;
}

int main(){
    std::vector<int> vec;
    int x=10;
    show(vec);
}

 

参考:http://en.cppreference.com/w/cpp/types/void_t

原理:using引入类型别名

#include <iostream>

template<int N,int M>
struct add{
    static constexpr int result = N + M;
};

template<int M>
using add8 = add<8,M>;

int main(){
    std::cout << add8<2>::result;
}

C++17起,支持了变长模板参数

template<typename...Args>
using void_t = void;

int f(void_t<int,char>){}
int f(void){}

int main(){
    f();
}

链接错误。原因在于void_t<int,char>等价于void,因此重定义了两次f();

 

posted @ 2018-03-19 15:08  thomas76  阅读(717)  评论(0编辑  收藏  举报