如何反转整数参数包?

stl / std 来自C++的库,因为我正在为一个只有可用的嵌入式操作系统编程。 gcc 4.4.4 用裸露的C++,所以,没有 std::tuple , std::forward , std::apply 或 std::anything_else

为了帮助理解元泛型生成的代码,我提供了一个用 clang 因为它可以显示生成的 template-meta-programming metaprogramming 代码。

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;
}

我只是在用 int array[] = {Sequence...} 做示范。实际使用的代码如下:

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
}

这就产生了这个吼声,对于像 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++ -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。而不是 int array[] = {5, 1, 0} ,应该是 int array[] = {0, 1, 5} .

我成功地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...

但是让我们假设我不能这样做,因为列表是从第三部分输入的,我没有控制权。如何反转参数包 <5, 1, 0> 到 <0, 1, 5>

在第一次尝试中,我尝试使用与生成整数列表相同的策略,但无法编译:

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;
}

当我尝试构建它时,出现了以下错误:

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>
                  ^

参考文献:

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?