使用 boost::adaptors 的 Matrix 类的 C++ 迭代器返回类型

Question

我正在尝试按行和列遍历自定义矩阵，并使用迭代器来遍历数据。

我在主文件中有一个工作代码，但我发现很难将其转换为 Matrix 类，因为我似乎无法确定 boost::begin 迭代器的返回类型/命令。这是工作代码：

#include <iostream>
#include <algorithm>                        //std::transform
#include "Matrix.h"                         //Matrix class
#include <boost/range/algorithm.hpp>        //boost::begin, boost::end
#include <boost/range/adaptor/strided.hpp>  //boost::adaptor::strided
#include <boost/range/adaptor/sliced.hpp>   //boost adaptor::slice


int main(int argc, char *argv[]){

    //creats a matrix with 5 rows and 3 colunms filled with ints from 0 to 14
    //stores the data internaly as a std::vector<T>
    Matrix<int, 5, 3> a = Matrix<int,5,3>(range<int>(15));

    //Matrix is accordingly overloaded, prints out the matrix 
    std::cout << a << std::endl;
    //returns an iterator to the second element and then traverses the vector by skipping 5 elments
    auto begin = boost::begin(boost::adaptors::stride(
                                  boost::adaptors::slice(a.as_vector(), 1, 15), 5));

    //returns an iterator to the end of the vector 
    auto end = boost::begin(boost::adaptors::stride(
                                boost::adaptors::slice(a.as_vector(), 1, 15), 5));

    //multiplies the second column times 2    
    std::transform(begin, end, begin, 
                   [](int i) { return 2*i;});

    //print result
    std::cout << a << std::endl;
}

运行程序返回：

$ ./main
[0,]   0   1   2 
[1,]   3   4   5 
[2,]   6   7   8 
[3,]   9  10  11 
[4,]  12  13  14 

[0,]   0   2   2 
[1,]   3   8   5 
[2,]   6  14   8 
[3,]   9  20  11 
[4,]  12  26  14 

正如我们所见，它有效。然而，矩阵类中开始/结束的实现导致了问题，特别是因为我需要将“auto”替换为，嗯，我真的不知道是什么。

class Matrix{
    //private data members
    //contuctors, functions
    //returns an iterator to the first element of vector
    typename std::vector<T>::iterator Begin(){
        return matrix.Begin();
    }

    typename std::vector<T>::iterator End(){
        return matrix.End();
    }

    //@ matrix: std::vector<T> that holds that data
    //@ fun - size(): returns the number of values stored in the matrix
    //@ fun - rows(): returns the number of rows of the matrix
    //@ param - i: column to iterate through
    typename ...return type in question... begin_col( std::size_t i ){
        return boost::begin(boost::adaptors::stride(
                                  boost::adaptors::slice( matrix, i, size() ), rows() ) );
            }

    typename ...return type in question... end_col( std::size_t i ){
        return boost::end(boost::adaptors::stride(
                                  boost::adaptors::slice( matrix, i, size() ), rows() ) );
            }
//many more overloaded operators and functions
//end of class
};

所以我希望 begin_col() 和 end_col() 的行为方式与两个 std::vector::iterator 几乎相同，仅适用于跨步切片。

矩阵类的其余部分可以在这里找到stackoverflow.com/questions/26282847/c-implementing-iterators-for-custom-matrix-class

最后编译代码： g++ -Wall -O3 -std=c++11 -o main main.cpp Matrix.cpp

在 ubuntu 14.04 上。

非常感谢任何 cmets。

谢谢

文森特

Answer 1

我花了一些时间，但这是一个有效的代码：

 ... certeris paribus

    auto begin_col( size_type i ){
        auto begin = boost::begin(boost::adaptors::stride(
                                boost::adaptors::slice(matrix.get_data(), i, size()), cols()) );
        return begin;
            }

    auto end_col( size_type i ){
        return boost::end(boost::adaptors::stride(
                                boost::adaptors::slice(matrix.get_data(), i, size()), cols()) );
            }

编译通过警告：

make -k 
g++ -Wall -O3 -std=c++11 -o main main.cpp Matrix.cpp Vector.cpp
In file included from main.cpp:5:0:
Matrix.h:116:33: warning: ‘begin_col’ function uses ‘auto’ type specifier without trailing return type [enabled by default]
 auto begin_col( size_type i ){

我不知道我是否应该关心，但我对结果很满意，因为：

//main.cpp ...
std::transform(a.begin_col(1), a.end_col(1), a.begin_col(1), []( int i ){ return 2*i; });
std::cout << a << std::endl;
//... 

返回：

$ ./main
[0,]   0   2   2 
[1,]   3   8   5 
[2,]   6  14   8 
[3,]   9  20  11 
[4,]  12  26  14 

--- End of test ---

就是这样。