Loki的Functor与可变模板 [英] Loki's Functor with variadic templates

问题描述

我有关于 Functor 库的实现 Loki 的问题。
我正在做一些更改，以使其与可变参数模板，而不是具有行和线的模板专业化。问题是，我试图使用typedef的可变参数模板，我不明白我的错误，这就是为什么我可以使用一些来自专家的帮助...

< a href =https://gist.github.com/athanase/7978090 =nofollow>这里是头文件。

I用一个简单的例子测试它：

  static void foo（）
 {
 std :: cout ;& foo !!! << std :: endl; 
} 
 int 
 main（int argc，const char ** argv）
 {
 Functor< void，void& static_func（foo）; 
 static_func（）; 
} 
  

这会给我这个错误

  /home/test/src/EntryPoint.cpp:237:17：error：没有匹配的调用'（Functor< void，void>）（）'
在/home/test/src/EntryPoint.cpp:231:0中包含的文件中：
 /home/test/include/FunctorTest.h:217:7：note：candidate is：
 / home /test/include/FunctorTest.h:292:16：note：Functor< R，TList> :: ResultType Functor< R，TList> :: operator（）（Functor< R，TList> :: MyList&& [with R = void; TList = {void}; Functor< R，TList> :: ResultType = void; functor< R，TList> :: MyList = variadic_typedef< void>] 
 /home/test/include/FunctorTest.h:292:16：note：candidate expects 1 argument，0 provided 
 / home / test / src / EntryPoint.cpp：在全局范围：
 /home/test/src/EntryPoint.cpp:234:1：warning：未使用的参数'argc'[-Wunused-parameter] 
 / home /test/src/EntryPoint.cpp:234:1：warning：unused参数'argv'[-Wunused-parameter] 
在/home/test/src/EntryPoint.cpp:231:0中包含的文件：
 /home/test/include/FunctorTest.h：在FunctorHandler< ParentFunctor，Fun>的实例化中：:: ResultType FunctorHandler< ParentFunctor，Fun> :: operator（）（FunctorHandler< ParentFunctor，Fun> :: MyList&& ;）[with ParentFunctor = Functor< void，void> ;; Fun = void（*）（）; FunctorHandler< ParentFunctor，Fun> :: ResultType = void; FuncterHandler< ParentFunctor，Fun> :: MyList = variadic_typedef< void>]'：
 /home/test/src/EntryPoint.cpp:247:1：从这里需要
 / home / test / include / FunctorTest.h：159：49：error：没有匹配的函数用于调用'forward（FunctorHandler< Functor< void，void> ;, void（*）（）> :: MyList&）'
 / home / test /include/FunctorTest.h:159:49：注意：候选项是：
在/usr/include/c++/4.7/bits/stl_pair.h:61:0中包含的文件中，
从/ usr /include/c++/4.7/utility:72，
从/home/jean/Lib/vitals/include/CLPair.h:28，
从/ home / jean / Lib / vitals / include / CLMap .h：27，
来自/home/jean/Lib/vitals/include/HTCmdLineParser.h:27，
来自/home/test/include/EntryPoint.h:23, 
从/home/test/src/EntryPoint.cpp:22：
 /usr/include/c++/4.7/bits/move.h:77:5：note：template< class _Tp> constexpr _Tp&& std :: forward（typename std :: remove_reference< _From> :: type&）
 /usr/include/c++/4.7/bits/move.h:77:5：note：template argument deduction / substitution failed： 
在/home/test/src/EntryPoint.cpp:231:0中包含的文件：
 /home/test/include/FunctorTest.h:159:49：note：无法转换'parms'（类型'FunctorHandler< Functor< void，void>，void（*）（）> :: MyList {aka variadic_typedef< void>}'）来键入'std :: remove_reference< convert_in_tuple< variadic_typedef< void& > > :: type& {aka convert_in_tuple< variadic_typedef< void> >&}&}'
  

解决方案

一个tuple函数直接使用std :: forward参数。
您需要这样的：

  / * 
 *指数< N + 1& ==>指数< 0,0,1,2，...，N> ==> Seq< 0,1,2，...，N> 
 * / 
 template< int ...> struct Seq {}; 
 
 template< int N，int ... S> 
 struct索引：索引< N-1，N-1，S ...> {}; 
 
 template< int ... S> 
 struct Indices< 0，S ...> {
 typedef Seq< S ...> SeqType; 
}; 
 
 template< typename R，typename F，typename ... A，int ... S> 
 inline R Apply（F f，const std :: tuple< A ...>& t，Seq< S ...>）{
 return（R）f get（S（t）...）; 
} 
 
 template< typename R，typename F，typename ... A> 
 inline R Apply（F f，const std :: tuple< A ...>& t）{
 return Apply< R>（f，t，typename Indices< sizeof ... A）> SeqType（））; 
} 
 
 template< typename R，typename T，typename F，typename ... A，int ... S> 
 inline R Apply（T * obj，F f，const std :: tuple< A ...>& t，Seq< S ...>）{
 return obj- * f）（std :: get S（t）...）; 
} 
 
 template< typename R，typename T，typename F，typename ... A> 
 inline R Apply（T * obj，F f，const std :: tuple< A ...>& t）{
 return Apply< R（obj，f，t，typename Indices< ; sizeof ...（A）> :: SeqType（））; 
} 
  

查看我的基于可变参数模板的Closure实现： https://github.com/idealvin/base/blob/master/closure.h p>

I have a question about the Functor implementation of the library Loki. I am doing some changes in order to make it work with variadic templates instead of having lines and lines of template specialization. The problem is that I am trying to use typedef for variadic template and I do not understand my error, that is why I could use some help from experts...

Here is the header file.

I tested it with a simple example:

static void foo()
{
    std::cout << "foo !!!" << std::endl;
}
int
main( int argc, const char** argv )
{
    Functor<void, void> static_func(foo);
    static_func();
}

Which gives me this error

/home/test/src/EntryPoint.cpp:237:17: error: no match for call to ‘(Functor<void, void>) ()’
    In file included from /home/test/src/EntryPoint.cpp:231:0:
    /home/test/include/FunctorTest.h:217:7: note: candidate is:
    /home/test/include/FunctorTest.h:292:16: note: Functor<R, TList>::ResultType Functor<R, TList>::operator()(Functor<R, TList>::MyList&&) const [with R = void; TList = {void}; Functor<R, TList>::ResultType = void; Functor<R, TList>::MyList = variadic_typedef<void>]
    /home/test/include/FunctorTest.h:292:16: note:   candidate expects 1 argument, 0 provided
    /home/test/src/EntryPoint.cpp: At global scope:
    /home/test/src/EntryPoint.cpp:234:1: warning: unused parameter ‘argc’ [-Wunused-parameter]
    /home/test/src/EntryPoint.cpp:234:1: warning: unused parameter ‘argv’ [-Wunused-parameter]
    In file included from /home/test/src/EntryPoint.cpp:231:0:
    /home/test/include/FunctorTest.h: In instantiation of ‘FunctorHandler<ParentFunctor, Fun>::ResultType FunctorHandler<ParentFunctor, Fun>::operator()(FunctorHandler<ParentFunctor, Fun>::MyList&&) [with ParentFunctor = Functor<void, void>; Fun = void (*)(); FunctorHandler<ParentFunctor, Fun>::ResultType = void; FunctorHandler<ParentFunctor, Fun>::MyList = variadic_typedef<void>]’:
    /home/test/src/EntryPoint.cpp:247:1:   required from here
    /home/test/include/FunctorTest.h:159:49: error: no matching function for call to ‘forward(FunctorHandler<Functor<void, void>, void (*)()>::MyList&)’
    /home/test/include/FunctorTest.h:159:49: note: candidates are:
    In file included from /usr/include/c++/4.7/bits/stl_pair.h:61:0,
                     from /usr/include/c++/4.7/utility:72,
                     from /home/jean/Lib/vitals/include/CLPair.h:28,
                     from /home/jean/Lib/vitals/include/CLMap.h:27,
                     from /home/jean/Lib/vitals/include/HTCmdLineParser.h:27,
                     from /home/test/include/EntryPoint.h:23,
                     from /home/test/src/EntryPoint.cpp:22:
    /usr/include/c++/4.7/bits/move.h:77:5: note: template<class _Tp> constexpr _Tp&& std::forward(typename std::remove_reference<_From>::type&)
    /usr/include/c++/4.7/bits/move.h:77:5: note:   template argument deduction/substitution failed:
    In file included from /home/test/src/EntryPoint.cpp:231:0:
    /home/test/include/FunctorTest.h:159:49: note:   cannot convert ‘parms’ (type ‘FunctorHandler<Functor<void, void>, void (*)()>::MyList {aka variadic_typedef<void>}’) to type ‘std::remove_reference<convert_in_tuple<variadic_typedef<void> > >::type& {aka convert_in_tuple<variadic_typedef<void> >&}’

解决方案

You can't apply a tuple to function parameters with std::forward directly. You need something like this:

/*
 * Indices<N + 1> ==> Indices<0, 0, 1, 2, ..., N> ==> Seq<0, 1, 2, ... , N>
 */
template<int...> struct Seq {};

template<int N, int ... S> 
struct Indices : Indices<N - 1, N - 1, S...> {};

template<int ... S> 
struct Indices<0, S...> { 
  typedef Seq<S...> SeqType; 
};

template<typename R, typename F, typename ... A, int ... S>
inline R Apply(F f, const std::tuple<A...>& t, Seq<S...>) {
  return (R) f(std::get<S>(t)...);
}

template<typename R, typename F, typename ... A>
inline R Apply(F f, const std::tuple<A...>& t) {
  return Apply<R>(f, t, typename Indices<sizeof...(A)>::SeqType());
}

template<typename R, typename T, typename F, typename ... A, int ... S>
inline R Apply(T* obj, F f, const std::tuple<A...>& t, Seq<S...>) {
  return (R) (obj->*f)(std::get<S>(t)...);
}

template<typename R, typename T, typename F, typename ... A>
inline R Apply(T* obj, F f, const std::tuple<A...>& t) {
  return Apply<R>(obj, f, t, typename Indices<sizeof...(A)>::SeqType());
}

See my implementation about Closure based on variadic templates: https://github.com/idealvin/base/blob/master/closure.h

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