有人可以帮我解决这个程序(使用指针转换矩阵)

时间:2013-07-02 23:03:31

标签: c++ pointers transpose

我制作的这个程序应该转换矩阵,无论它的大小是多少。然而,它不像它应该工作,我不知道为什么,我确实得到它的输出没有编译错误,但{1,2,3},{4,5,6}的输出是{ 1,0,0},{0,0,4},这对我来说毫无意义。我曾多次在纸上写过“记忆快照”,但却找不到我所遗漏的东西,此时我真的需要另外一套眼睛,谢谢。

#include "stdafx.h"

#include <fstream>

#include <iostream>

#include <iomanip>

#include <string>

#include <cmath>

#include <vector>

using namespace std;

void squaretranspose(int &M, int &MT, int ROWS, int COLS);

int main(void)
{
    int M[2][3]={{1,2,3},{4,5,6}};
    int MT[3][2]={0};
    int ROWS(2),COLS(3);

int i,j;
cout << " The entries of the original matrix " << endl;
for(i=0;i<=ROWS-1;i++)
{
    for(j=0;j<=COLS-1;j++)
{
cout<<M[i][j]<<"\t";
}
cout << endl;
}
squaretranspose(M[0][0],MT[0][0],ROWS,COLS);
cout << " The entries of the transposed non-square matrix " << endl;
for(i=0;i<=COLS-1;i++)
{
    cout << endl;
    for(j=0;j<=ROWS-1;j++)
{
cout<<MT[i][j]<<"\t";
}
}
system ("PAUSE");
return 0;
}
void squaretranspose (int &M, int &MT, int ROWS, int COLS)
{
    // declare pointers to change the input matrice's values
    int *ptr,*ptrT;
// declare indices for a row by row process
    int i,j;
// declare placeholder 2d vectors for swapping the I,j, entries to ,j,i entries  
    vector < vector<int>> temp(ROWS,COLS);
    vector < vector<int>> tempT(COLS,COLS);
    vector < vector<int>> temp_T(ROWS,ROWS);
// set the pointers to point to the first entry of the input and output matrices
    ptr = &M;
    ptrT = &MT;

// if rows=cols we want to use 2d vector temp
    if (ROWS=COLS)
    {
// store all of the input matrice's values in the 2d vector "temp"
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {

// set the i,j th entry of the 2d vector "temp" equal to the value currently pointed to by the pointer
        temp[i][j]=*ptr;
// increment the pointer to the address of the next entry of the input matrix unless we are on the last entry
               if ((i!=ROWS-1)&&(j!=COLS-1))
               { 
               ptr++;
               }
        }
        }
    }
// reset pointer address to first entry
    ptr=&M;
// the for loop for swapping the j,i entries of the 2d vector "temp" with the i,j entries of the input matrix
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {
// if j is not equal to i swap the value pointed to by the pointer (the i,j entry of the input matrix) for the j,ith entries value of the 2d vector "temp"
        if (j!=i)
            {
            *ptr=temp[j][i];
            }
        // increment the pointer if it is not on the last entry
               if ((i!=ROWS-1)&&(j!=COLS-1))
               { 
               ptr++;
               }
        }
    }*/


// if ROWS<COLS we want to have 2d vector tempT
if (ROWS<COLS)
{
// store all of the input matrice's values in the 2d vector "tempT"
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {
// set the j,ith entry of the 2d vector "temp" equal to the value currently pointed to by the pointer
        tempT[j][i]=*ptr;
// increment the pointer to the address of the next entry of the input matrix
        if (((i!=(ROWS-1))&&(j!=(COLS-1))))
        {
            ptr++;
        }
        }
    }
    ptr=&M;
// transport the entries of tempT into the output matrix MT
    for(i=0;i<=COLS-1;i++)
    {
        for(j=0;j<=ROWS-1;j++)
        {
        *ptrT=tempT[i][j];
// increment the pointer
        if (((i!=ROWS-1)&&(j!=COLS-1)))
        {
            ptrT++;
        }
        }
    }
}
ptrT=&MT;

// if ROWS>COLS we want to use the 2d vector temp_T
if (ROWS>COLS)
{
// store all of the input matrice's values in the 2d vector "temp_T"
    for(i=0;i<=ROWS-1;i++)
    {
        for(j=0;j<=COLS-1;j++)
        {

// set the j,i th entry of the 2d vector "temp" equal to the value currently pointed to by the pointer
        temp_T[j][i]=*ptr;
// increment the pointer
        if (((i!=ROWS-1)&&(j!=COLS-1)))
        {
            ptrT++;
        }
    }

// the for loop for swapping the j,i entries of the 2d vector "temp" with the i,j entries of the input matrix
    for(i=0;i<=COLS-1;i++)
    {
        for(j=0;j<=ROWS-1;j++)
        {
// if j is not equal to i swap the value pointed to by the pointer (the i,j entry of the input matrix) for the j,ith entries value of the 2d vector "temp"
            if (j!=i)
            {
            *ptrT=temp_T[j][i];
            }
// increment the pointer
        if (((i!=ROWS-1)&&(j!=COLS-1)))
        {
            ptrT++;
        }
        }
    }
}
return;
}

//所以它就是这样,如果COLS非常小并且由于中间的2d方形向量而说ROWS非常大,它不是非常有效,但对于其他一切它应该是非常好的,如果它正常工作。

2 个答案:

答案 0 :(得分:0)

使用DIY索引:ptrT[ index(j,i,COLS,ROWS) ] = ptr[ index(i,j,ROWS,COLS) ]。现在你所要做的就是编写将2D indeces转换为1D indeces的函数。

答案 1 :(得分:0)

您的代码存在几个主要问题,这一行:

if (ROWS=COLS)

可能意味着:

if (ROWS == COLS)

第一种情况是,您将COLS的值分配给ROWS,在第二种情况下,您将检查它们是否相等。在您应该使用for的所有<=循环中,您正在使用<,否则您将访问阵列边界之外的其他循环。

除此之外,代码太复杂了,转置功能应该非常简单,这是一种可能的方法:

template <int n, int m>
void squaretranspose( int a[n][m], int b[m][n])
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            b[j][i] = a[i][j];
        }
    }
} 

但转换矩阵的最快最简单的方法是反转坐标,因此您无需访问(i,j)即可访问(j,i)。另一方面,如果性能是您的主要关注点,那么这个previous thread也很好地涵盖了我的解决方案,并且我的解决方案可能适合您。