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        如何使用可变参数模板制作通用的 Lua 函数包装器?

        时间:2023-09-27
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                  本文介绍了如何使用可变参数模板制作通用的 Lua 函数包装器?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  对于我当前的项目,我一直在为 Lua 包装器编写大量 C/C++.其中很多是简单的 setter 和 getter,所以我设法编写了一些模板来轻松生成这些,如下所示:

                  For my current project, I've been writing a lot of C/C++ to Lua wrappers. A large number of these are simple setters and getters, so I managed to write some templates that make it easy to generate these, like so:

                  //        Class       Return      Field
                  template <typename T, typename U, U T::*Member>
                  int luaU_get(lua_State* L)
                  {
                      T* obj = luaW_check<T>(L, 1);
                      luaU_push<U>(L, obj->*Member);
                      return 1;
                  }
                  
                  static luaL_reg Foo_Table[] =
                  {
                      ...
                      // Now I can just use this generic template to avoid
                      // writing simple getter functions
                      { "getbar", luaU_get<Foo, Bar, &Foo::bar> }, 
                      ...
                  };
                  

                  我也想为任意函数的简单函数包装器做类似的事情.例如,能够做到这一点就好了:

                  I would like to do something similar for simple function wrappers for arbitrary functions too. For example, it would be nice to be able to do this:

                  template <typename T, typename U, U (T::*Func)(), typename... Args>
                  int luaU_func(lua_State* L)
                  {
                       // ...?
                  }
                  
                  static luaL_reg Foo_Table[] =
                  {
                      ...
                      { "baz", luaU_func<Foo, int, &Foo::baz, int, float> }, 
                      ...
                  };
                  

                  这个想法是模板在编译时实际上是这样的:

                  The idea is that the template effectively turn out to be this when compiled:

                  int luaU_func(lua_State* L)
                  {
                       luaU_push<int>(L, luaW_check<Foo>(L, 1)->baz(luaU_check<int>(L, 2), luaU_check<float>(L, 3)));
                       return 1;
                  }
                  

                  我试过只使用 ... 扩展器,对我来说问题是整数索引值映射到正确的参数.我想不出让它们正常工作的方法.这种事情有可能吗?

                  I've tried just using the ... expander, the problem for me is the integer index values to map to the proper arguments. I can't think of a way to get them working right. Is such a thing even possible?

                  (这里已经有了一些魔法;我为 lua_push 和 lua_check 之类的东西写了一些模板化的包装器.所有这些现有的包装器都可以找到 这里)

                  (there is a little bit of magic going on here already; I wrote some templated wrappers for things like lua_push and lua_check. All of those existing wrappers can be found here)

                  推荐答案

                  诀窍是通过部分特化包含包装函数的类模板来利用模板参数推导:

                  The trick is to exploit template argument deduction by partially specializing a class template that contains the wrapper function:

                  // Lua API dummies ...
                  
                  struct lua_State {};
                  
                  template<class T> void luaU_push(lua_State*,T);
                  template<class T> T* luaW_check(lua_State*,int);
                  template<class T> T luaU_check(lua_State*,int);
                  
                  
                  // metaprogramming for creating indices ...
                  
                  template<int...Ints>
                  struct int_pack {};
                  
                  template<int Begin, int Count, int...Tail>
                  struct make_int_range_type {
                      typedef typename make_int_range_type<Begin,Count-1,Begin+Count-1,Tail...>::type type;
                  };
                  
                  template<int Begin, int...Tail>
                  struct make_int_range_type<Begin,0,Tail...> {
                      typedef int_pack<Tail...> type;
                  };
                  
                  template<int Begin, int Count>
                  inline typename make_int_range_type<Begin,Count>::type
                  make_int_range()
                  { return typename make_int_range_type<Begin,Count>::type(); }
                  
                  
                  // the actual wrapper ...
                  
                  template<class MemFunPtrType, MemFunPtrType PMF>
                  struct lua_mem_func_wrapper;
                  
                  template<class Clazz, class ReturnType, class...Args, ReturnType(Clazz::*PMF)(Args...)>
                  struct lua_mem_func_wrapper<ReturnType(Clazz::*)(Args...),PMF> {
                      static int doit(lua_State* L) {
                          return doit_impl(L,make_int_range<2,sizeof...(Args)>());
                      }
                  private:
                      template<int...Indices>
                      static int doit_impl(lua_State* L, int_pack<Indices...>) {
                          luaU_push<ReturnType>(L,
                              (luaW_check<Clazz>(L, 1)->*PMF)(
                                  luaU_check<Args>(L, Indices)...
                              )
                          );
                          return 1;
                      }
                  };
                  
                  #define GET_MEM_FUN_WRAPPER(...) &lua_mem_func_wrapper<decltype(__VA_ARGS__),__VA_ARGS__>::doit
                  
                  
                  // testing ...
                  
                  struct foo {
                      int baz(int, float);
                  };
                  
                  void test() {
                      auto* ptr = GET_MEM_FUN_WRAPPER(&foo::baz);
                  }
                  

                  此代码使用选项 -c --std=c++0x 在 G++ 4.6.1 下编译.要查看它是否真的符合您的要求,请对其进行测试...

                  This code compiles under G++ 4.6.1 using the options -c --std=c++0x. To see whether it really does what you want, please test it ...

                  这篇关于如何使用可变参数模板制作通用的 Lua 函数包装器?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跟版网!

                  上一篇:从Lua调用函数时如何处理C++异常? 下一篇:在 C/C++ 中引用 Lua 函数

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