C中的矩阵乘法与来自文件的输入

Question

我已经编写了矩阵乘法程序，但是我想对其进行修改。首先，我想将矩阵改为first [a] [b]而不是10，并且从文件中读取矩阵的维数。我是否需要根据使用malloc的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从文件中存储矩阵的维数。建议我需要哪些更改？我不打开文件，只是将stdin重定向到文件。我无法通过文件获得输入？

修改后的代码作为

#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; 
} 

Answer 1

变化的声明：

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

后scanf("%d%d", &m, &n);：

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

BEF矿石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)); 

免费声明：

更换（在程序结束时）（如果不出口权后的程序总是必填项）：

} 
    return 0; 
} 

附：

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; 
} 

的另一种方法：分配的存储器中的连续块

声明：

int *first, *second, *multiply; 

malloc的：（无for循环）

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

用法：

• 变化first[c][k]到first[c*m+k]
• 变化second[k][d]到second[k*p+d]
• 变化multiply[c][d]到first[c*m+d]

free的：（无for循环）

• free(first);
• free(second);
• free(multiply);