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,或者我可以采取一些最大值,但是那会导致大量内存的浪费 我需要将矩阵的维度从文件存储在数组中吗?建议我需要做什么改变?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;
}
最佳答案
更改声明:
int i, m, n, p, q, c, d, k, sum = 0;
int **first, **second, **multiply;
scanf("%d%d", &m, &n);
之后:
first = malloc(m*sizeof(int*));
for (i = 0; i < m; i++)
first[i] = malloc(n*sizeof(int));
在printf("输入第二个矩阵的元素\n");
之前:
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) p>
替换:
}
return 0;
}
与:
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:分配连续的内存块
声明:
int *first, *second, *multiply;
malloc
的:(无 for 循环)
first = malloc(m*n*sizeof(int));
第二个 = malloc(p*q*sizeof(int));
乘法 = malloc(m*q*sizeof(int));
用法:
- 将
first[c][k]
更改为first[c*m+k]
- 将
秒[k][d]
更改为秒[k*p+d]
- 将
multiply[c][d]
更改为first[c*m+d]
免费
的:(无for循环)
免费(第一);
免费(第二);
自由(乘法);
关于c - 使用文件输入的 C 矩阵乘法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14924229/