Question

我有一个任务，我必须使矩阵乘法程序更有效所以我写了一个叫做乘法矩阵的方法但是在我实际上在循环中进行矩阵乘法之后，最终的产品矩阵都是零但是，如果我在循环中检查它不是零

int** multiply_matrices(int** matrix1, int** matrix2, int m1, int n1, int m2, int n2) {
if(n1 != m2) {
    printf("error\n");
    return NULL;
}

int i = 0;
int j = 0;
int k = 0;
int** product = (int**) malloc(m1 * sizeof(int*));



for(i = 0; i<m1; i++){
    product[i] = (int*) malloc(n2 * sizeof(int));
    for(j = 0; j<n1; j++){
        product[i][j] = 0;
    }
}
*
for(i = 0; i < m1; i ++) {
    product[i] = (int*) malloc(n2 * sizeof(int));
    int chg = 0;

     while(j<n2){


        if(k == n1 ){
            //chg = 0;
            k = 0;
            //printf("%d\n", j);
            j++;
        }

             product[i][j]  += matrix1[i][k] * matrix2[k][j];

            printf("%d \n", product[i][j]);
        k++;
        chg = 1;

    }
}

return product;

}

Answer 1

Two errors I found:

You aren't resetting j = 0 in the outer multiplication loop. This leads to it only completing one row, if it enters the inner loop at all, as its value remains equal to n1 after the allocation loop. This is most likely why it is returning all zeros for you.

Second, you should reset k in the outer loop as well. When the condition j<n2 is broken, i increments but k remains the same value as when the last loop ended.

I also do not know what the point of the chg variable is as is it doing nothing.

And, as TrustworthySystems said, only allocate memory once or you will have leaks. Here is the correct function:

int** multiply_matrices(int** matrix1, int** matrix2, int m1, int n1, int m2, int n2) {
if(n1 != m2) {
    printf("error\n");
    return NULL;
}

int i = 0;
int j = 0;
int k = 0;
int** product = (int**) malloc(m1 * sizeof(int*));



for(i = 0; i<m1; i++){
    product[i] = (int*) malloc(n2 * sizeof(int));
    for(j = 0; j<n1; j++){
        product[i][j] = 0;
    }
}

for(i = 0; i < m1; i++) {
    int chg = 0;
    j = 0;
    k = 0;

    while(j<n2){
        if(k == n1 ){
            k = 0;
            j++;
        }
        product[i][j]  += matrix1[i][k] * matrix2[k][j];
        k++;
        chg = 1;

    }
}
return product;
}

c语言循环后的矩阵乘法错误

1 个答案: