我为Strassen矩阵乘法编写了以下代码。我知道它很大，但你不需要经历整个事情。我的问题是在编译期间，参数为[] [num]，b [] [num]和c [] [num]的Strassen函数不具有固定值num。这是我出错的地方。我需要在main中输入num的输入，这就是为什么它不能全局赋值。我怎样才能解决这个问题？我的代码：
#include <stdio.h>
int num;
void strassen(int a[][num], int b[][num], int c[][num], int size) {
int p1[size/2][size/2], p2[size/2][size/2], p3[size/2][size/2], p4[size/2][size/2], p5[size/2][size/2], p6[size/2][size/2], p7[size/2][size/2];
int temp1[size/2][size/2], temp2[size/2][size/2];
int q1, q2, q3, q4, q5, q6, q7, i, j;
if(size >= 2) { //give recursive calls
//p1
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i][j] + a[i + size / 2][j + size / 2];
}
}
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i][j] + b[i + size / 2][j + size / 2];
}
}
num = size / 2;
strassen(temp1, temp2, p1, size / 2);
//p2
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i + size / 2][j] + a[i + size / 2][j + size / 2];
}
}
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i][j];
}
}
num = size / 2;
strassen(temp1, temp2, p2, size / 2);
//p3
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i][j];
}
}
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i][j + size / 2] - b[i + size / 2][j + size / 2];
}
}
num = size / 2;
strassen(temp1, temp2, p3, size / 2);
//p4
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i + size / 2][j + size / 2];
}
}
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i + size / 2][j] - b[i][j];
}
}
num = size / 2;
strassen(temp1, temp2, p4, size / 2);
//p5
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i][j] + a[i][j + size / 2];
}
}
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i + size / 2][j + size / 2];
}
}
num = size / 2;
strassen(temp1, temp2, p5, size / 2);
//p6
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i + size / 2][j] - a[i][j];
}
}num = size / 2;
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i][j] + b[i][j + size / 2];
}
}
num = size / 2;
strassen(temp1, temp2, p6, size / 2);
//p7
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp1[i][j] = a[i][j + size / 2] - a[i + size / 2][j + size / 2];
}
}
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
temp2[i][j] = b[i + size / 2][j] + b[i + size / 2][j + size / 2];
}
}
num = size / 2;
strassen(temp1, temp2, p7, size / 2);
//c11
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
c[i][j] = p1[i][j] + p4[i][j] - p5[i][j] + p7[i][j];
}
}
//c12
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
c[i][j + size / 2] = p3[i][j] + p5[i][j];
}
}
//c21
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
c[i + size / 2][j] = p2[i][j] + p4[i][j];
}
}
//c22
for(i = 0; i < size / 2; i++) {
for(j = 0; j < size / 2; j++) {
c[i + size / 2][j + size / 2] = p1[i][j] + p3[i][j] - p2[i][j] + p6[i][j];
}
}
}
else if(size == 1) {
c[0][0] = a[0][0] * b[0][0];
}
}
int padding(int num) {
int original_num = num, lower_power = 0, i, actual_num = 1;
if(num == 1)
return 1;
while(num > 1) {
lower_power++;
num /= 2;
}
for(i = 0; i < lower_power; i++) {
actual_num *= 2;
}
if(actual_num == original_num)
return original_num;
else
return actual_num * 2;
}
int main() {
int i, j, temp;
printf("Enter the size of nxn matrix:\n");
scanf("%d", &num);
temp = num;
if(num <= 0)
return 0;
num = padding(num);
int a[num][num], b[num][num], c[num][num];
printf("Enter matrix a:\n"); //accept inputs for a and b from the user
for(i = 0; i < temp; i++) {
for(j = 0; j < temp; j++) {
scanf("%d", &a[i][j]);
}
for(j = temp; j < num; j++) {
a[i][j] = 0;
}
}
for(i = temp; i < num; i++)
for(j = 0; j < num; j++)
a[i][j] = 0;
printf("\nEnter matrix b:\n");
for(i = 0; i < temp; i++) {
for(j = 0; j < temp; j++) {
scanf("%d", &b[i][j]);
}
for(j = temp; j < num; j++) {
b[i][j] = 0;
}
}
for(i = temp; i < num; i++)
for(j = 0; j < num; j++)
b[i][j] = 0;
printf("Matrix a:\n"); //printing the actual matrices for strassen's multiplication
for(i = 0; i < num; i++) {
for(j = 0; j < num; j++) {
printf("%d ", a[i][j]);
}
printf("\n");
}
printf("\nMatrix b:\n");
for(i = 0; i < num; i++) {
for(j = 0; j < num; j++) {
printf("%d ", b[i][j]);
}
printf("\n");
}
strassen(a, b, c, num);
printf("\nMatrix c is:\n");
for(i = 0; i < temp; i++) {
for(j = 0; j < temp; j++) {
printf("%d ", c[i][j]);
}
printf("\n");
}
return 0;
}
答案 0 :(得分：1)
您可以（甚至应该）使用int **代替，并传递num作为参数。作为简化示例，以下是如何计算动态大小的数组的总和：
int sum(int *in, int len) {
int out = 0;
for(int i = 0; i < len; i++)
out += in[i];
return out;
}