Boost.MPL转换状态？

Question

我有以下mpl序列

boost::mpl::vector_c<std::size_t, 0, 1, 2, 0, 1, 0>

我需要根据以下算法(运行时版本)对其进行转换:

i=0
output_sequence=[]
for k in (0,...,len(input_sequence)-1):
if input_sequence[k] == 0:
    output_sequence.append(i)
    i=i+1
else:
    output_sequence.append(-1)

我的结果应该是:

boost::mpl::vector_c<std::size_t, 0, -1, -1, 1, -1, 2>

我可以想象至少有两种方法可以在运行时使用std :: transform或std :: accumulate来实现,但是不知道如何在编译时使用mpl实现相同的结果.对我来说,主要的问题是以某种方式存储状态'i'(当前发现的零数)以及输出序列.

非常感谢你!

添加

基于HighCommander4的回答,我想与大家分享所采用的方法.实际上,可以使用boost::mpl::fold自定义元函数.与HighCommander4相比,唯一的概念变化是将当前可用的自由索引打包为增长输出序列的第一个元素.它可以在类型计算结束时使用删除boost::mpl::pop_font.

#include <boost/mpl/size_t.hpp>
#include <boost/mpl/equal_to.hpp>

#include <boost/mpl/fold.hpp>

#include <boost/mpl/front.hpp>
#include <boost/mpl/push_front.hpp>
#include <boost/mpl/push_back.hpp>
#include <boost/mpl/pop_front.hpp>

#include <boost/mpl/vector_c.hpp>
#include <boost/mpl/if.hpp>
#include <boost/mpl/at.hpp>

#include <boost/mpl/print.hpp>

namespace mpl=boost::mpl;

struct assign_index
{
                enum { dim_dynamic, dim_static };
                template<typename Output, typename Index,
                         typename Enable = typename mpl::if_<
                                             mpl::equal_to<Index, mpl::size_t<1> >,
                                             mpl::size_t<dim_dynamic>,
                                             mpl::size_t<dim_static> >::type>
                struct apply 
                {
                                typedef typename mpl::push_back<Output, mpl::size_t<-1> >::type type;
                };


                template<typename Output, typename Index>
                struct apply <Output, Index, mpl::size_t<dim_dynamic> >
                {
                                typedef typename mpl::front<Output>::type current;
                                typedef typename mpl::next<current>::type next;
                                typedef typename mpl::push_back<Output, current>::type append_type;
                                typedef typename mpl::push_front<
                                                typename mpl::pop_front<append_type>::type,next
                                                >::type type;
                };              
};

template<class input_sequence>
struct map_indices
{
                // The first element of state0 keeps track of the current index count.
                typedef mpl::vector_c<std::size_t, 0> state0;
                typedef typename mpl::fold<input_sequence, state0, assign_index>::type output_sequence;

                // Remove the first element from the final output sequence
                typedef typename mpl::pop_front<output_sequence>::type type;
};

int main (int argc, const char** argv)
{
                typedef mpl::vector_c<std::size_t, 1, 2, 3, 1, 2, 1> input_sequence;
                typedef map_indices<input_sequence>::type output_sequence;
                int i;
                i=mpl::print<mpl::at_c<output_sequence, 0>::type>();
                i=mpl::print<mpl::at_c<output_sequence, 1>::type>();
                i=mpl::print<mpl::at_c<output_sequence, 2>::type>();
                i=mpl::print<mpl::at_c<output_sequence, 3>::type>();
                i=mpl::print<mpl::at_c<output_sequence, 4>::type>();
                i=mpl::print<mpl::at_c<output_sequence, 5>::type>();
                return 0;
}

PS:在此实现中,输入序列中的"1"被转换为递增索引,而大于1的值被转换为std :: size_t(-1).输入序列中不允许使用零(在别处检查).

PS2:我已经检查过,确实没有设施boost::mpl可以将任意mpl序列(作为算法结果获得的序列)转换回来boost::mpl::vector.因此,有必要为此编写代码...我最终调整了代码

<boost/fusion/container/vector/detail/as_vector.hpp>

"规范化"类型计算结果.

再次感谢你!

Answer 1

模板元编程确实需要一些人习惯.

注意循环如何转换为递归,如果语句情况转换为模板特化,并"声明"为其他模板参数:

#include <cstddef>
#include <boost/mpl/print.hpp>
#include <boost/mpl/vector_c.hpp>
#include <boost/mpl/push_back.hpp>
#include <boost/mpl/pop_front.hpp>
#include <boost/mpl/front.hpp>
#include <boost/mpl/size.hpp>

using boost::mpl::vector_c;
using boost::mpl::push_back;
using boost::mpl::pop_front;
using boost::mpl::front;
using boost::mpl::int_;
using boost::mpl::size;

typedef vector_c<std::size_t, 0, 1, 2, 0, 1, 0> input_sequence;

template <int i, int current>
struct next_i
{
    static const int value = i;
};

template <int i>
struct next_i<i, 0>
{
    static const int value = i + 1;
};

template <int i, int current>
struct next_output
{
    static const int value = -1;
};

template <int i>
struct next_output<i, 0>
{
    static const int value = i;
};

template <typename rest_of_input, int rest_of_input_size, int i, typename output_so_far>
struct loop
{
    typedef typename front<rest_of_input>::type current_t;
    static const int current = current_t::value;
    static const int next_i_ = next_i<i, current>::value;
    static const int next_output_ = next_output<i, current>::value; 

    typedef typename loop<
            typename pop_front<rest_of_input>::type,
            rest_of_input_size - 1,
            next_i_,
            typename push_back<output_so_far, int_<next_output_> >::type 
    >::type type;
};

template <typename rest_of_input, int i, typename output_so_far>
struct loop<rest_of_input, 0, i, output_so_far>
{
    typedef output_so_far type;
};

template <typename input>
struct algorithm
{
    typedef typename size<input>::type size_;
    typedef typename loop<input, size_::value, 0, vector_c<std::size_t> >::type type;
};

typedef algorithm<input_sequence>::type output_sequence;

int main()
{
    boost::mpl::print<output_sequence>();
}

这可能写得更好(也许是mpl::fold).