13

考虑以下功能:

template<typename... List> 
inline unsigned int myFunction(const List&... list)
{
    return /* SOMETHING */; 
}

/* SOMETHING */为了返回sizeof所有参数的总和,最简单的方法是什么?

例如myFunction(int, char, double) = 4+1+8 = 13

4

6 回答 6

16
unsigned myFunction() {return 0;}

template <typename Head, typename... Tail>
unsigned myFunction(const Head & head, const Tail &... tail) {
    return sizeof head + myFunction(tail...);
}
于 2012-10-01T02:14:31.357 回答
16

在 C++17 中,使用折叠表达式:

template<typename... List> 
inline constexpr unsigned int myFunction(const List&... list)
{
    return (0 + ... + sizeof(List));
}
于 2016-03-18T18:38:21.010 回答
4

根据此评论和对该问题的以下评论,您可以使用它(注意:完全未经测试)

std::initializer_list<std::size_t> sizeList = {sizeof(List)...}; //sizeList should be std::initializer_list, according to the comments I linked to
return std::accumulate(sizeList.begin(), sizeList.end(), 0);
于 2012-10-01T02:16:38.847 回答
3

晚了两年,但保证由编译器计算一个替代解决方案(如果你不介意不同的语法):

template < typename ... Types >
struct SizeOf;

template < typename TFirst >
struct SizeOf < TFirst >
{
    static constexpr auto Value = (sizeof(TFirst));
};

template < typename TFirst, typename ... TRemaining >
struct SizeOf < TFirst, TRemaining ... >
{
    static constexpr auto Value = (sizeof(TFirst) + SizeOf<TRemaining...>::Value);
};

用作constexpr std::size_t size = SizeOf<int, char, double>::Value; // 4 + 1 + 8 = 13

于 2016-03-18T18:18:01.133 回答
0

这是一个模板方式:

#include <iostream>

template<typename T, typename ...Ts>
class PPackSizeOf
{
  public:
  static const unsigned int size = sizeof(T) + PPackSizeOf<Ts...>::size;
};


template<typename T>
class PPackSizeOf<T>
{
  public:
  static const unsigned int size = sizeof(T);
};

template<typename ...Ts>
class FixedSizeBlock
{
   private:
      char block[PPackSizeOf<Ts...>::size];
   public:

};

int main( )
{
  FixedSizeBlock<char,long> b;
  std::cout << sizeof(b) << std::endl; 
  return 0;
}
于 2016-08-19T15:44:29.273 回答
-3

我刚刚发现:

template<typename... List> 
inline unsigned int myFunction(const List&... list)
{
    return sizeof(std::make_tuple(list...)); 
}

但 :

1)我是否保证所有编译器的结果总是相同的?

2) make_tuple 会在编译时生成和开销吗?

于 2012-10-01T02:20:06.400 回答