如何反转整数参数包? [英] How to reverse an integer parameter pack?

问题描述

可悲的是，我不能使用 stl 的任何问题 / std 库的问题，因为我正在为嵌入式操作系统，仅具有可用的 gcc 使用裸露的C ++，因此没有std::tuple，std::forward，std::apply或std::anything_else.

Sadly, I cannot use any of stl/std libraries from C++, because I am programming for a embedded Operating System which only has available gcc 4.4.4 with bare C++, so, no std::tuple, std::forward, std::apply or std::anything_else.

为帮助您理解生成的元通用代码，我将展示一个使用 clang ，因为它可以选择向我们显示生成的 template-meta-programming /

To help understand meta generic generated code, I am presenting a minimal example code compiled with clang because it has a option to show us the generated template-meta-programming/metaprogramming code.

这个问题只是出于好奇，因为我可以以正确的顺序创建它，而不是以错误的顺序生成整数参数包.这是我用来按错误顺序生成整数打包程序包的方法:

This question is just for curiosity because instead of generating the integer parameter pack in the wrong order, I can just create it on a correct order. This is what I use to generate my integer packager pack on the wrong order:

template<int ...>
struct MetaSequenceOfIntegers { };

template<int AccumulatedSize, typename Tn, int... GeneratedSequence>
struct GeneratorOfIntegerSequence;

template<int AccumulatedSize, typename Grouper, typename Head, typename... Tail, int... GeneratedSequence>
struct GeneratorOfIntegerSequence< AccumulatedSize, Grouper( Head, Tail... ), GeneratedSequence... >
{
    typedef typename GeneratorOfIntegerSequence
            < AccumulatedSize + sizeof(Head), Grouper( Tail... ), AccumulatedSize, GeneratedSequence...
            >::type type;
};

template<int AccumulatedSize, typename Grouper, int... GeneratedSequence>
struct GeneratorOfIntegerSequence<AccumulatedSize, Grouper(), GeneratedSequence...>
{
  typedef MetaSequenceOfIntegers<GeneratedSequence...> type;
};

template<int ...Sequence>
void intergers_sequencer_generator(MetaSequenceOfIntegers<Sequence...>) {
    int array[] = {Sequence...};
}

int main(int argc, char const *argv[]) {
    intergers_sequencer_generator( GeneratorOfIntegerSequence< 0, int(char, int, char) >::type() );
    return 0;
}

我仅使用int array[] = {Sequence...}进行演示.实际使用的代码是这样的:

I am only using int array[] = {Sequence...} for demonstration. The real code used is like this:

template<typename ReturnType, typename... Tn>
class Closure
{ 
    // ... other code
    template<int ...Sequence>
    ReturnType _run(MetaSequenceOfIntegers<Sequence...>) {
        return _function_entry_pointer( get_nth_function_argument_on_address<Sequence, Tn>()... );
    }
    // ... other code
}

为以下输入生成此波纹管，例如create_functor( &function1, 'a', 10, 'b' ):

Which generates this bellow, for an input like create_functor( &function1, 'a', 10, 'b' ):

template <int ...Sequence> char _run(MetaSequenceOfIntegers<Sequence...>);
template<> char _run<<5, 1, 0>>(MetaSequenceOfIntegers<5, 1, 0>)     {
    return this->_function_entry_pointer(
            this->get_nth_function_argument_on_address<5, const char *>(), 
            this->get_nth_function_argument_on_address<1, const char *>(), 
            this->get_nth_function_argument_on_address<0, char>()
        );
}
// and much more

我们可以使用clang 查看生成的代码:

We can see the generated code using clang:

$ clang++ -Xclang -ast-print -fsyntax-only generator.cpp > expanded.cpp

template <int ...> struct MetaSequenceOfIntegers {
};
template<> struct MetaSequenceOfIntegers<<5, 1, 0>> {
};
template <int AccumulatedSize, typename Tn, int ...GeneratedSequence> struct GeneratorOfIntegerSequence
template<> struct GeneratorOfIntegerSequence<0, int (char, int, char), <>> {
    typedef typename GeneratorOfIntegerSequence<0 + sizeof(char), int (int, char), 0>::type type;
}
template<> struct GeneratorOfIntegerSequence<1, int (int, char), <0>> {
    typedef typename GeneratorOfIntegerSequence<1 + sizeof(int), int (char), 1, 0>::type type;
}
template<> struct GeneratorOfIntegerSequence<5, int (char), <1, 0>> {
    typedef typename GeneratorOfIntegerSequence<5 + sizeof(char), int (), 5, 1, 0>::type type;
}
template<> struct GeneratorOfIntegerSequence<6, int (), <5, 1, 0>> {
    typedef MetaSequenceOfIntegers<5, 1, 0> type;
};
template <int AccumulatedSize, typename Grouper, typename Head, typename ...Tail, int ...GeneratedSequence> struct GeneratorOfIntegerSequence<AccumulatedSize, type-parameter-0-1 (type-parameter-0-2, type-parameter-0-3...), <GeneratedSequence...>> {
    typedef typename GeneratorOfIntegerSequence<AccumulatedSize + sizeof(Head), Grouper (Tail...), AccumulatedSize, GeneratedSequence...>::type type;
};
template <int AccumulatedSize, typename Grouper, int ...GeneratedSequence> struct GeneratorOfIntegerSequence<AccumulatedSize, type-parameter-0-1 (), <GeneratedSequence...>> {
    typedef MetaSequenceOfIntegers<GeneratedSequence...> type;
};
template <int ...Sequence> void intergers_sequencer_generator(MetaSequenceOfIntegers<Sequence...>) {
    int array[] = {Sequence...};
}
template<> void intergers_sequencer_generator<<5, 1, 0>>(MetaSequenceOfIntegers<5, 1, 0>) {
    int array[] = {5, 1, 0};
}
int main(int argc, const char *argv[]) {
    intergers_sequencer_generator(GeneratorOfIntegerSequence<0, int (char, int, char)>::type());
    return 0;
}

生成元编程列表的顺序是我需要的相反顺序.代替int array[] = {5, 1, 0}，它应该是int array[] = {0, 1, 5}.

The meta programming list is generated is on reversed order I need it to be. Instead of int array[] = {5, 1, 0}, it should be int array[] = {0, 1, 5}.

我只是在示例代码中更改了这一行，才设法以正确的顺序生成了列表:

I managed to generated the list on the correct order just changing this line on in example code:

< AccumulatedSize + sizeof(Head), Grouper( Tail... ), GeneratedSequence..., AccumulatedSize
// to -->
< AccumulatedSize + sizeof(Head), Grouper( Tail... ), AccumulatedSize, GeneratedSequence...

但是让我们假设我不能这样做，因为该列表是从我没有控制权的第三部分输入的.如何在不使用任何标准库函数的情况下将参数包<5, 1, 0>反向转换为<0, 1, 5>?

But let us supposed I could not do that because the list is input from a third part which I do not have control. How can I reverse the parameter pack <5, 1, 0> to <0, 1, 5> without using any std library functions?

在我的第一次尝试中，我尝试使用与生成整数列表相同的策略，但是无法对其进行编译:

In my first attempt, I tried to use the same strategy as the one I did to generate the integer list, but I could not make it to compile:

template<int ...>
struct MetaSequenceOfIntegers { };

template<int AccumulatedSize, typename Tn, int... GeneratedSequence>
struct GeneratorOfIntegerSequence;

template<int AccumulatedSize, typename Grouper, typename Head, typename... Tail, int... GeneratedSequence>
struct GeneratorOfIntegerSequence< AccumulatedSize, Grouper( Head, Tail... ), GeneratedSequence... >
{
    typedef typename GeneratorOfIntegerSequence
            < AccumulatedSize + sizeof(Head), Grouper( Tail... ), AccumulatedSize, GeneratedSequence...
            >::type type;
};

template<int AccumulatedSize, typename Grouper, int... GeneratedSequence>
struct GeneratorOfIntegerSequence<AccumulatedSize, Grouper(), GeneratedSequence...>
{
  typedef MetaSequenceOfIntegers<GeneratedSequence...> type;
};

// The new code starts here
template<int ...>
struct MetaSequenceReversed { };

template<typename Tn, int... GeneratedSequence>
struct ReversorOfIntegerSequence;

template<typename Grouper, int Head, int... Tail, int... GeneratedSequence>
struct ReversorOfIntegerSequence< Grouper( Head, Tail... ), GeneratedSequence... >
{
    typedef typename ReversorOfIntegerSequence
            < Grouper( Tail... ), GeneratedSequence...
            >::type type;
};

template<typename Grouper, int... GeneratedSequence>
struct ReversorOfIntegerSequence<Grouper(), GeneratedSequence...>
{
  typedef MetaSequenceReversed<GeneratedSequence...> type;
};

template<int ...ReversedSequence>
void intergers_sequencer_reversor(MetaSequenceReversed<ReversedSequence...>) {
    int reversed_array[] = {ReversedSequence...};
}

template<int ...Sequence>
void intergers_sequencer_generator(MetaSequenceOfIntegers<Sequence...>) {
    int array[] = {Sequence...};
    intergers_sequencer_reversor( ReversorOfIntegerSequence< int(Sequence...) >::type() );
}

int main(int argc, char const *argv[])
{
    intergers_sequencer_generator( GeneratorOfIntegerSequence< 0, int(char, int, char) >::type() );
    return 0;
}

当我尝试构建它时，出现此错误:

When I try to build this, I got this error:

generator.cpp:29:35: error: template argument for template type parameter must be a type
struct ReversorOfIntegerSequence< Grouper( Head, Tail... ), GeneratedSequence... >
                                  ^~~~~~~~~~~~~~~~~~~~~~~~
generator.cpp:25:19: note: template parameter is declared here
template<typename Tn, int... GeneratedSequence>
                  ^
generator.cpp:50:62: error: template argument for template type parameter must be a type
    intergers_sequencer_reversor( ReversorOfIntegerSequence< int(Sequence...) >::type() );
                                                             ^~~~~~~~~~~~~~~~
generator.cpp:25:19: note: template parameter is declared here
template<typename Tn, int... GeneratedSequence>
                  ^

---

参考文献:

---

References:

1. 可变参数模板，参数包及其在参数列表中讨论的模糊性
2. 一个元组以调用匹配的函数指针
3. 可以用C ++看到模板实例化的代码吗?编译器
4. 使用可变参数模板构建函数参数
5. 如何反转顺序可变参数模板函数的参数集?

1. Variadic templates, parameter pack and its discussed ambiguity in a parameter list
2. "unpacking" a tuple to call a matching function pointer
3. Can we see the template instantiated code by C++ compiler
4. Build function parameters with variadic templates
5. How to reverse the order of arguments of a variadic template function?

推荐答案

如何在不使用任何标准库函数的情况下将参数pack <5, 1, 0>转换为<0, 1, 5>?

不确定要确切地使用什么功能，但是...对我来说似乎很容易.

Not sure to understand what do you exactly can use but... it seems easy to me.

给出如下的帮助结构

template <typename, typename>
struct RS_helper;

template <int ... As, int B0, int ... Bs>
struct RS_helper<MetaSequenceOfIntegers<As...>,
                 MetaSequenceOfIntegers<B0, Bs...>>
   : RS_helper<MetaSequenceOfIntegers<B0, As...>,
               MetaSequenceOfIntegers<Bs...>>
 { };

template <typename T>
struct RS_helper<T, MetaSequenceOfIntegers<>>
 { using type = T; };

还原结构可以很简单

template <int ... Is>
struct RevertSequence
   : RS_helper<MetaSequenceOfIntegers<>, MetaSequenceOfIntegers<Is...>>
 { };

我想反向函数可能有用

template <int ... Is>
constexpr typename RevertSequence<Is...>::type
   revertSequenceFunction (MetaSequenceOfIntegers<Is...> const &)
 { return {}; }

我建议对原始代码进行修改，并增加反向序列(也使用std::cout打印序列，但显然可以将其删除).

I propose a modified version of your original code with addition of reversed sequence (using also std::cout to print the sequences, but you can remove it, obviously).

#include <iostream>

template <int ...>
struct MetaSequenceOfIntegers
 { };

template <int AccumulatedSize, typename Tn, int ... GeneratedSequence>
struct GeneratorOfIntegerSequence;

template <int AccumulatedSize, typename Grouper, typename Head,
          typename ... Tail, int ... GeneratedSequence>
struct GeneratorOfIntegerSequence<AccumulatedSize, Grouper(Head, Tail...),
                                  GeneratedSequence... >
 { typedef typename GeneratorOfIntegerSequence
            <AccumulatedSize+sizeof(Head), Grouper(Tail...),
             AccumulatedSize, GeneratedSequence...>::type type; };

template <int AccumulatedSize, typename Grouper, int ... GeneratedSequence>
struct GeneratorOfIntegerSequence<AccumulatedSize, Grouper(),
                                  GeneratedSequence...>
 { typedef MetaSequenceOfIntegers<GeneratedSequence...> type; };

template <int ... Sequence>
void intergers_sequencer_generator(MetaSequenceOfIntegers<Sequence...>)
 {
   using unused = int[];

   (void)unused { 0, (std::cout << Sequence << ' ', 0)... };

   std::cout << std::endl;
 }


template <typename, typename>
struct RS_helper;

template <int ... As, int B0, int ... Bs>
struct RS_helper<MetaSequenceOfIntegers<As...>,
                 MetaSequenceOfIntegers<B0, Bs...>>
   : RS_helper<MetaSequenceOfIntegers<B0, As...>,
               MetaSequenceOfIntegers<Bs...>>
 { };

template <typename T>
struct RS_helper<T, MetaSequenceOfIntegers<>>
 { using type = T; };

template <int ... Is>
struct RevertSequence
   : RS_helper<MetaSequenceOfIntegers<>, MetaSequenceOfIntegers<Is...>>
 { };

template <int ... Is>
constexpr typename RevertSequence<Is...>::type
   revertSequenceFunction (MetaSequenceOfIntegers<Is...> const &)
 { return {}; }

int main ()
 {
   intergers_sequencer_generator(
      GeneratorOfIntegerSequence<0, int(char, int, char)>::type());

   intergers_sequencer_generator(
      revertSequenceFunction(
         GeneratorOfIntegerSequence<0, int(char, int, char)>::type()));
 }

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