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c++ - 谁能帮我解决这个程序的问题(它使用指针转置矩阵)

转载 作者:太空宇宙 更新时间:2023-11-04 14:05:01 26 4
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我有一个我制作的程序,它应该转置一个矩阵,无论它的大小是多少。然而,它并没有像预期的那样工作,我不知道为什么,我确实从它那里得到了输出,没有编译错误,但是 {1,2,3},{4,5,6} 的输出是 { 1,0,0},{0,0,4},这对我来说完全没有意义。我已经多次在纸上浏览“内存快照”,但就是找不到我遗漏的东西,此时我真的只需要另一双眼睛,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;
}

//就是这样,如果 COLS 非常小并且说 ROWS 非常大,因为中间的 2d 平方 vector ,它不是很有效,但对于其他一切它应该是相当好的,如果它工作正常。

最佳答案

您的代码有几个主要问题,这一行:

if (ROWS=COLS)

可能意味着:

if (ROWS == COLS)

第一种情况你将赋值为COLSROWS在第二种情况下,您将检查它们是否相等。在你所有的for你正在使用的循环 <=什么时候应该使用 <否则您将访问数组边界之外的一个。

除此之外,代码太复杂了,转置函数应该非常简单,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];
}
}
}

但是转置矩阵的最快和最简单的方法是反转坐标,而不是访问 (i,j)你访问(j,i) .另一方面,如果性能是您的主要关注点,那么这个 previous thread我的解决方案也很好地涵盖了这个主题,并且我的解决方案与其他可能适合您的解决方案一样。

关于c++ - 谁能帮我解决这个程序的问题(它使用指针转置矩阵),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17437218/

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