C ++ 0x：迭代一个带有函数的元组 [英] C++0x: iterating through a tuple with a function

问题描述

我有一个名为 _push 的函数，它可以处理不同的参数，包括元组，并且应该返回推送元素的数量。

例如， _push（5）应该在堆栈上按下'5'（ _push（std :: make_tuple（） （5，hello））应该按'5'和'hello'并返回2。

_push（5，hello）因为我有时使用 _push（foo（）） foo（）返回一个元组。

无论如何我不能使它与元组一起工作： / p>

  template< typename ... Args，int N = sizeof ...（Args）> 
 int _push（const std :: tuple< Args ...>& t，typename std :: enable_if<（N> = 1）> :: type * = nullptr）{
 return _push< Args ...，N-1>（t）+ _push（std :: get< N-1（t））; 
} 
 
 template< typename ... Args，int N = sizeof ...（Args）> 
 int _push（const std :: tuple< Args ...>& t，typename std :: enable_if<（N == 0）> :: type * = nullptr）{
 return 0; 
} 
  

假设您要推送 tuple< int ，bool> 。这是我预期的工作方式：

• _push< {int，bool}，2>

• _push< {int，bool}，1>


• _push< {int，bool}，0> ul>
  
  

  然而，使用g ++ 4.5（我支持可变模板的唯一编译器），我得到一个关于 _push< Args ...，N-1> ;（t）（第3行）说它找不到匹配的函数调用（没有任何进一步的细节）。我尝试没有...但我得到另一个错误，说明参数包不扩展。

  
  
  

  如何解决这个问题？

  
  
  PS：我知道你可以使用模板struct（这实际上是我之前做的），但我想知道如何做一个

  
  
  

  >解决方案

  我解决了一些黑客的问题。这里是代码：

    template< typename ... Args，int N = sizeof ...（Args）> 
   int _push（const std :: tuple< Args ...>& t，std :: integral_constant< int，N> * = nullptr，typename std :: enable_if& ; :: type * = nullptr）{
   return _push（t，static_cast< std :: integral_constant< int，N-1>（nullptr））+ _push（std :: get< N-1> （t））; 
  } 
   template< typename ... Args，int N = sizeof ...（Args）> 
   int _push（const std :: tuple< Args ...>& t，std :: integral_constant< int，N> * = nullptr，typename std :: enable_if<（N == 0）> :: type * = nullptr）{
   return 0; 
  } 
    

  如果找到更好的办法，请不要犹豫发布

  
  

  I have a function named _push which can handle different parameters, including tuples, and is supposed to return the number of pushed elements.

  For example, _push(5) should push '5' on the stack (the stack of lua) and return 1 (because one value was pushed), while _push(std::make_tuple(5, "hello")) should push '5' and 'hello' and return 2.

  I can't simply replace it by _push(5, "hello") because I sometimes use _push(foo()) and I want to allow foo() to return a tuple.

  Anyway I can't manage to make it work with tuples:

  template<typename... Args, int N = sizeof...(Args)>
  int _push(const std::tuple<Args...>& t, typename std::enable_if<(N >= 1)>::type* = nullptr) {
   return _push<Args...,N-1>(t) + _push(std::get<N-1>(t));
  }
  
  template<typename... Args, int N = sizeof...(Args)>
  int _push(const std::tuple<Args...>& t, typename std::enable_if<(N == 0)>::type* = nullptr) {
   return 0;
  }
  

  Let's say you want to push a tuple<int,bool>. This is how I expect it to work:

  • _push<{int,bool}, 2> is called (first definition)
  • _push<{int,bool}, 1> is called (first definition)
  • _push<{int,bool}, 0> is called (second definition)

  However with g++ 4.5 (the only compiler I have which supports variadic templates), I get an error concerning _push<Args...,N-1>(t) (line 3) saying that it couldn't find a matching function to call (without any further detail). I tried without the "..." but I get another error saying that the parameters pack is not expanded.

  How can I fix this?

  PS: I know that you can do this using a template struct (this is in fact what I was doing before), but I'd like to know how to do it with a function

  PS 2: PS2 is solved, thanks GMan

  解决方案

  I solved the problem with some hacks. Here is the code:

  template<typename... Args, int N = sizeof...(Args)>
  int _push(const std::tuple<Args...>& t, std::integral_constant<int,N>* = nullptr, typename std::enable_if<(N >= 1)>::type* = nullptr) {
      return _push(t, static_cast<std::integral_constant<int,N-1>*>(nullptr)) + _push(std::get<N-1>(t));
  }
  template<typename... Args, int N = sizeof...(Args)>
  int _push(const std::tuple<Args...>& t, std::integral_constant<int,N>* = nullptr, typename std::enable_if<(N == 0)>::type* = nullptr) {
      return 0;
  }
  

  Don't hesitate to post if you find a better way

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