如何反转整数参数包

遗憾的是，我不能使用C++中的任何stl/std库，因为我是为嵌入式操作系统编程的，该操作系统只有可用的带有裸C++的gcc4.4.4，因此没有std::tuple、std::forward、std::apply或std::anything_else。

为了帮助理解元通用生成代码，我将展示一个用clang编译的最小示例代码，因为它可以向我们展示生成的模板元编程/元编程代码。

这个问题只是为了好奇，因为我可以按照正确的顺序创建它，而不是按照错误的顺序生成整数参数包。这是我用来在错误的顺序上生成整数包装包的东西：

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;
}

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

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
}

对于像create_functor( &function1, 'a', 10, 'b' ):这样的输入，它生成以下内容

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看到生成的代码：

$ 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;
}

元编程列表的生成顺序与我需要的相反。它应该是int array[] = {5, 1, 0}，而不是int array[] = {0, 1, 5}。

我设法按照正确的顺序生成了列表，只是在示例代码中更改了这一行：

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

但我们假设我不能这样做，因为列表是从第三部分输入的，我无法控制。如何在不使用任何std库函数的情况下将参数包<5, 1, 0>反转为<0, 1, 5>？

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，但我无法编译：

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;
}

当我试图构建这个时，我得到了这个错误：

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>
^

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