变换元组类型 [英] Transform tuple type

问题描述

我想要一个覆盖这种类型的元函数：

  std :: tuple< T0，T1，...，TN& 
  

进入：

  std :: tuple< 
 std :: function< T0（std :: tuple< T0，T1，...>，std :: tuple< T0，T1，...>），
 std： ：function< T1（std :: tuple< T0，T1，...>，std :: tuple< T0，T1，...>），
 ...，
 std :: function< TN（...）> 
> 
  

看起来像这样可以通过 transform ，但我想有一个元组类型，而不是一个类型的向量。 （它不必使用MPL实际上，但我想它会更短？）

背景：目前我使用完全通用的类型，并依赖于所有地狱破如果使用错误，但我想计算 TupleOfFunctions 以获得正确的错误。

  template< class TupleOfValues，class TupleOfFunctions> 
 void f（TupleOfValues v，TupleOfFunctions fun）
  

解决方案

p>以下内容？

 模板< typename T> struct transform; 
 template< typename ... T> 
 struct transform< std :: tuple< T ...>> {
 typedef std :: tuple< std :: function< T，std :: tuple< T ...> ;, std :: tuple< T ...>> ...&类型; 
}; 
  

So I'm new to boost MPL, and I don't know how to use it with standard types.

I want a metafunction that coverts this type:

std::tuple<T0, T1, ..., TN>

Into this:

std::tuple<
  std::function<T0(std::tuple<T0, T1, ...>, std::tuple<T0, T1, ...>)>,
  std::function<T1(std::tuple<T0, T1, ...>, std::tuple<T0, T1, ...>)>,
  ...,
  std::function<TN(...)>
>

and it seems like this could be done with transform, but I want to have a tuple-type, not a vector of types. (It doesn't have to use MPL actually, but I guess it would be shorter?)

Background: currently I use totally generic types and rely on all hell breaking loose if used wrong, but I want to calculate the TupleOfFunctions to get a proper error.

template<class TupleOfValues, class TupleOfFunctions>
void f(TupleOfValues v, TupleOfFunctions fun)

解决方案

How about the following?

template<typename T> struct transform;
template<typename ...T>
struct transform<std::tuple<T...>> {
  typedef std::tuple<std::function<T, std::tuple<T...>, std::tuple<T...>>...> type;
};

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