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我已经编写了一个矩阵乘法程序，但我想修改它。首先，我想将矩阵更改为 first[a][b] 而不是 10 并从文件中读取矩阵的维度。我是否需要使用 malloc 根据矩阵的维度动态分配内存，或者我可以取一些最大值，但是会导致大量内存的浪费我是否需要将矩阵的维度存储在文件中的数组中。建议我需要哪些更改？我没有打开文件，只是将标准输入重定向到文件。我无法通过文件获取输入？？

修改后的代码如下

#include <stdio.h>

int main() 
{
 int m, n, p, q, c, d, k, sum = 0;
  int **first, **second, **multiply;
  printf("Enter the number of rows and columns of first matrix\n");
  scanf("%d%d", &m, &n);
  first = malloc(m*sizeof(int*));
   for (int i =0;i <m; i++)
first[i] =malloc(n*sizeof(int));
    second = malloc(p*sizeof(int*));
     for(int i=0;i<p;i++)
second[i] = malloc(q*sizeof(int));
multiply = malloc(m*sizeof(int));
for (int i=0;i<q;i++)
multiply[i] = malloc(q*sizeof(int));
printf("Enter the elements of first matrix\n");
    for (  c = 0 ; c < m ; c++ )
    for ( d = 0 ; d < n ; d++ )
      scanf("%d", &first[c][d]);
    printf("Enter the number of rows and columns of second matrix\n");
  scanf("%d%d", &p, &q);
  if ( n != p )
    printf(
  "Matrices with entered orders can't be multiplied with each other.\n");
  else {
 printf("Enter the elements of second matrix\n");
 for ( c = 0 ; c < p ; c++ )
  for ( d = 0 ; d < q ; d++ )
    scanf("%d", &second[c][d]);
 for ( c = 0 ; c < m ; c++ ) {
  for ( d = 0 ; d < q ; d++ ) {
    for ( k = 0 ; k < p ; k++ ) {
      sum = sum + first[c][k]*second[k][d];
    }
    multiply[c][d] = sum;
    sum = 0;
  }
  }
 printf("Product of entered matrices:-\n");
 for ( c = 0 ; c < m ; c++ ) {
  for ( d = 0 ; d < q ; d++ )
    printf("%d\t", multiply[c][d]);
  printf("\n");
 }
for (int i = 0; i < p; i++)
  free(second[i]);
free(second);
for (int i = 0; i < q; i++)
  free(multiply[i]);
  free(multiply);
}
for (int i = 0; i < m; i++)
free(first[i]);
 free(first);
return 0;
 }
4

1 回答 1

4

变更声明：

int i, m, n, p, q, c, d, k, sum = 0;
int **first, **second, **multiply;

之后scanf("%d%d", &m, &n);：

first = malloc(m*sizeof(int*));
for (i = 0; i < m; i++)
  first[i] = malloc(n*sizeof(int));

之前printf("Enter the elements of second matrix\n");：

second = malloc(p*sizeof(int*));
for (i = 0; i < p; i++)
  second[i] = malloc(q*sizeof(int));

multiply = malloc(m*sizeof(int*));
for (i = 0; i < q; i++)
  multiply[i] = malloc(q*sizeof(int));

自由语句：（在程序结束时）（如果程序没有立即退出，则始终需要）

代替：

  }
  return 0;
}

和：

    for (i = 0; i < p; i++)
      free(second[i]);
    free(second);
    for (i = 0; i < q; i++)
      free(multiply[i]);
    free(multiply);
  }
  for (i = 0; i < m; i++)
    free(first[i]);
  free(first);
  return 0;
}

另一种方法：分配一个连续的内存块

声明：

int *first, *second, *multiply;

malloc's: (没有for循环)

  • first = malloc(m*n*sizeof(int));
  • second = malloc(p*q*sizeof(int));
  • multiply = malloc(m*q*sizeof(int));

用法：

  • 更改first[c][k]为first[c*m+k]
  • 更改second[k][d]为second[k*p+d]
  • 更改multiply[c][d]为first[c*m+d]

free's: (没有for循环)

  • free(first);
  • free(second);
  • free(multiply);
于 2013-02-17T18:11:26.827 回答