类型列表的运行时类型开关作为开关而不是嵌套的 if? [英] Runtime typeswitch for typelists as a switch instead of a nested if's?

问题描述

这是来自 TTL:

////////////////////////////////////////////////////////////
//  run-time type switch
template <typename L, int N = 0, bool Stop=(N==length<L>::value) > struct type_switch;

template <typename L, int N, bool Stop>
  struct type_switch
  {
    template< typename F >
      void operator()( size_t i, F& f )
      {
        if( i == N )
        {
          f.operator()<typename impl::get<L,N>::type>();
        }
        else
        {
          type_switch<L, N+1> next;
          next(i, f);
        }
      }
  };

它用于在 TypeList 上进行类型切换.问题是——他们通过一系列嵌套的 if 来做到这一点.有没有办法将这种类型切换作为单个选择语句来代替?

It's used for typeswitching on a TypeList. Question is -- they are doing this via a series of nested if's. Is there a way to do this type switch as a single select statement instead?

谢谢！

推荐答案

你需要预处理器来生成一个大的switch.您需要 get<> 来进行无操作越界查找.检查编译器输出以确保未使用的情况不会产生输出，如果您关心的话；根据需要进行调整；v).

You'll need the preprocessor to generate a big switch. You'll need get<> to no-op out-of-bound lookups. Check the compiler output to be sure unused cases produce no output, if you care; adjust as necessary ;v) .

如果您想精通这类事情，请查看 Boost 预处理器库……

Check out the Boost Preprocessor Library if you care to get good at this sort of thing…

template <typename L>
  struct type_switch
  {
    template< typename F >
      void operator()( size_t i, F& f )
      {
        switch ( i ) {
         #define CASE_N( N ) 
         case (N): return f.operator()<typename impl::get<L,N>::type>();
         CASE_N(0)
         CASE_N(1)
         CASE_N(2)
         CASE_N(3) // ad nauseam.
      }
  };

这篇关于类型列表的运行时类型开关作为开关而不是嵌套的 if?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！