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目前， boost::fusion::for_each 迭代单个序列的元素。我正在尝试创建一个以类似方式工作但具有许多序列的函数，并将迭代序列之间的所有可能组合。

例如，如果我有三个序列 S1、S2、S3，我想创建一个这样的函子

struct my_functor {

template <class x, class y, class z>
void operator()(x& el1, y& el2, z& el3) {...}
}

然后打电话

for_each(s1, s2, s3, my_functor()) // applies the functor to all combinations of elements of s1, s2, s3

其中 s1, s2, s3 是 S1, S2, S3 的实例。

我首先为一般情况（任意数量的序列）编写代码，但发现它太难了。所以我决定从两个序列开始，然后从那里开始。当我有两个序列（为简单起见假设 fusion::vectors ）时，我已经设法完成了，如下所示：

//for_each.hpp

#include <boost/fusion/include/mpl.hpp>
#include <boost/fusion/include/at_c.hpp>
#include <boost/fusion/include/vector.hpp>
#include <boost/fusion/include/back.hpp>
#include <boost/mpl/size.hpp>

template <class Seq1, class Seq2, int i1, int i2, class F>                                     
struct my_call {                                                           

  static void apply(Seq1& seq1, Seq2& seq2, F& f) {                                        
    f(boost::fusion::at_c<i1>(seq1), boost::fusion::at_c<i2>(seq2)); // apply functor for a given pair of ints             
    my_call<Seq1, Seq2, i1, i2+1, F>::apply(seq1, seq2, f); // increase second int by 1 and apply functor again            
  }                                                                
};                                                                 

// terminal condition for 2nd sequence                                                 
template <class Seq1, class Seq2, int i1, class F>                                         
struct my_call<Seq1, Seq2, i1, boost::mpl::size<Seq2>::type::value - 1, F> {                               

  static void apply(Seq1& seq1, Seq2& seq2, F& f) {                                        
    f(boost::fusion::at_c<i1>(seq1), boost::fusion::back(seq2));                                   
    my_call<Seq1, Seq2, i1+1, 0, F>::apply(seq1, seq2, f); // reset 2nd int and increase 1st by 1                  
  }                                                                
};                                                                 

// terminal condition for both sequences                                               
template <class Seq1, class Seq2, class F>                                             
struct my_call<Seq1, Seq2, boost::mpl::size<Seq2>::type::value - 1, boost::mpl::size<Seq2>::type::value - 1, F> {          

  static void apply(Seq1& seq1, Seq2& seq2, F& f) {                                        
    f(boost::fusion::back(seq1), boost::fusion::back(seq2));                                       
  }                                                                
};                                                                 


// the actual function                                                         
template <class Seq1, class Seq2, class F>                                             
void for_each(Seq1& seq1, Seq2& seq2, F& f) {                                              
  my_call<Seq1, Seq2, 0, 0, F>::apply(seq1, seq2, f);                                          
}

和主要作为

//main.cpp
#include "for_each.hpp"
#include <iostream>

struct myf {
  template <class X, class Y>
  void operator()(X& x, Y& y) {
    std::cout << x + y << std::endl;
  }
};

int main() {
  boost::fusion::vector<int, double> x(1, 2.5);
  boost::fusion::vector<double, int> y(2, 5);
  myf F;
  for_each(x, y, F);
  return 0;
}

我的主要（没有双关语）问题是概括上述内容以使其适用于任意数量的序列。非常欢迎任何建议！谢谢

4

1 回答 1

1

您可以制作序列序列并将它们传递给调用者函数。

int main()
{
    boost::fusion::vector<int, double> x(1, 2.5);
    boost::fusion::vector<double, int> y(2, 5);
    boost::fusion::vector<double, double, double> z(10, 20, 30);

    CallFunc(myf(), boost::fusion::make_vector(x, y));
    CallFunc(myf(), boost::fusion::make_vector(x, y, z));
}

从每个序列的CallFunc元素生成笛卡尔积，然后将它们传递给给定的函子。

#include <iostream>

#include <boost/fusion/container/vector.hpp>
#include <boost/fusion/container/generation/make_vector.hpp>
#include <boost/fusion/algorithm/iteration/for_each.hpp>
#include <boost/fusion/include/empty.hpp>
#include <boost/fusion/include/pop_front.hpp>
#include <boost/fusion/include/front.hpp>
#include <boost/fusion/include/push_back.hpp>
#include <boost/fusion/include/invoke.hpp>

struct myf {
    typedef void result_type;

    template <class X, class Y>
    void operator()(X x, Y y) {
        std::cout << x + y << std::endl;
    }
    template <class X, class Y, class Z>
    void operator()(X x, Y y, Z z) {
        std::cout << x + y + z << std::endl;
    }
};

template<class Stop> struct CallFuncOuter;

template<class Func, class Tail, class CallTuple>
struct CallFuncInner
{
    CallFuncInner(Func &f, Tail &seq, CallTuple & args)
        : f(f)
        , tail(seq)
        , args(args)
    {
    }

    template<class HeadArg>
    void operator()(HeadArg & head_arg) const
    {
        CallFuncOuter<boost::fusion::result_of::empty<Tail>::type>()
            (f, tail, boost::fusion::push_back(args, head_arg));
    }

    Func &f;
    Tail &tail;
    CallTuple &args;
};

template<class Func, class Tail, class CallTuple>
CallFuncInner<Func, Tail, CallTuple> MakeCallFuncInner(Func &f, Tail &tail, CallTuple &arg)
{
    return CallFuncInner<Func, Tail, CallTuple>(f, tail, arg);
}

template<class Stop>
struct CallFuncOuter
{
    template<class Func, class SeqOfSeq, class CallTuple>
    void operator()(Func &f, SeqOfSeq & seq, CallTuple & args) const
    {
        boost::fusion::for_each(boost::fusion::front(seq),
            MakeCallFuncInner(
                f,
                boost::fusion::pop_front(seq),
                args));
    }
};

template<>
struct CallFuncOuter<boost::mpl::true_>
{
    template<class Func, class SeqOfSeq, class CallTuple>
    void operator()(Func &f, SeqOfSeq & seq, CallTuple & args) const
    {
        boost::fusion::invoke(f, args);
    }
};

template<class Func, class SeqOfSeq>
void CallFunc(Func &f, SeqOfSeq & seq)
{
    CallFuncOuter<boost::fusion::result_of::empty<SeqOfSeq>::type>()
        (f, seq, boost::fusion::vector<>());
}
于 2014-07-11T13:07:05.830 回答