將std :: tuple <T...>轉換爲T

Question

因此，我給出了一個std::tuple<T...>，我想創建一個接受T...的函數指針，目前這是我所得到的;

template<typename... Arguments> 
using FunctionPointer = void (*)(Arguments...); 

using FunctionPtr = FunctionPointer<typename std::tuple_element<0, V>::type, 
            typename std::tuple_element<1, V>::type, 
            typename std::tuple_element<2, V>::type>; 

不過，我似乎無法找到一個方法來做到這一點，沒有從0, ..., tuple_size<V>::value手動輸入每一個指標。 FunctionPtr被定義在一個上下文中，其中V=std::tuple<T...>（也已經有一個可變模板（因此我不能直接通過T...））

我想我需要生成一些索引列表，並做一些黑魔法..

Answer 1

這裏是一個可能的解決方案：

#include <tuple> 

// This is what you already have... 
template<typename... Arguments> 
using FunctionPointer = void (*)(Arguments...); 

// Some new machinery the end user does not need to no about 
namespace detail 
{ 
    template<typename> 
    struct from_tuple { }; 

    template<typename... Ts> 
    struct from_tuple<std::tuple<Ts...>> 
    { 
     using FunctionPtr = FunctionPointer<Ts...>; 
    }; 
} 

//================================================================= 
// This is how your original alias template ends up being rewritten 
//================================================================= 
template<typename T> 
using FunctionPtr = typename detail::from_tuple<T>::FunctionPtr; 

這裏是你將如何使用它：

// Some function to test if the alias template works correctly 
void foo(int, double, bool) { } 

int main() 
{ 
    // Given a tuple type... 
    using my_tuple = std::tuple<int, double, bool>; 

    // Retrieve the associated function pointer type... 
    using my_fxn_ptr = FunctionPtr<my_tuple>; // <== This should be what you want 

    // And verify the function pointer type is correct! 
    my_fxn_ptr ptr = &foo; 
} 

Answer 2

一個簡單的特質英里GHT做的伎倆：

#include <tuple> 

template <typename> struct tuple_to_function; 

template <typename ...Args> 
struct tuple_to_function<std::tuple<Args...>> 
{ 
    typedef void (*type)(Args...); 
}; 

用法：

typedef std::tuple<Foo, Bar, int> myTuple; 

tuple_to_function<myTuple>::type fp; // is a void (*)(Foo, Bar, int)