所以我的尝试是创建一个程序，自动从 .txt 文件中获取两个矩阵的大小并将它们相乘。我可以制作具有给定大小的程序，所以我本身只有计算列和行的问题。
输入类似于（MxN 矩阵）：
1 2 3 4
1 2 3 4
1 2 3 4
具体来说，这是我到目前为止的程序（我认为代码的开头是相关的）：
#include <stdio.h>
#include <stdlib.h>
struct mat1
{
int cols;
int rows;
};
struct mat2
{
int cols;
int rows;
};
struct mat1 dim1(const char* file)
{
struct mat1 m1;
int rows = 0;
int cols = 0;
char c;
FILE *f = fopen(file, "r+");
while((c = fgetc(f) != EOF))
{
if(c != '\n' && rows == 0)
{
cols++;
}
else if(c == '\n')
rows++;
}
rows++;
return m1;
}
struct mat2 dim2(const char* file)
{
struct mat2 m2;
int rows = 0;
int cols = 0;
char c;
FILE *f = fopen(file, "r+");
while((c = fgetc(f) != EOF))
{
if(c != '\n' && rows == 0)
{
cols++;
}
else if(c == '\n')
rows++;
}
rows++;
return m2;
}
double* alloc_matrix(int cols, int rows) {
double* m = (double*)malloc(cols * rows * sizeof(double));
if (m == 0) {
printf("Memory allocation error.\n");
exit(-1);
}
return m;
}
void read_matrix(FILE* f, double* m, int cols, int rows) {
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
fscanf(f, "%lf", &m[i * cols + j]);
}
}
}
void multiplication(double* m1, double* m2, double* m3, int cols, int rows) {
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
m3[i * cols +j]=0;
for(int k = 0; k < cols; k++) {
m3[i * cols +j]+=m1[i * cols +k]*m2[k * cols +j];
}
}
}
}
void write_matrix(double* m, int cols, int rows) {
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
printf("%f ", m[i * cols + j]);
}
printf("\n");
}
}
int main(int argc, char* argv[])
{
char* matrix1 = argv[1];
char* matrix2 = argv[2];
if (argc < 3) {
printf("Not enough arguments.\n");
exit(-1);
}
struct mat1 m1 = dim1(matrix1);
struct mat2 m2 = dim2(matrix2);
printf(" %d %d \n", m1.cols, m1.rows);
printf(" %d %d \n", m2.cols, m2.rows);
int c1 = m1.cols;
int r1 = m1.rows;
int c2 = m2.cols;
int r2 = m2.rows;
if (r1!=c2)
{
printf("Matrixes are not suitable for multiplication. \n");
exit(-1);
}
double* mtx1 = alloc_matrix(c1, r1);
double* mtx2 = alloc_matrix(c2, r2);
FILE* f1 = fopen(matrix1, "r");
if (f1 == 0)
{
printf("Cannot open file %s.", argv[1]);
exit(-1);
}
FILE* f2 = fopen(matrix2, "r");
if (f1 == 0)
{
printf("Cannot open file %s.", argv[1]);
exit(-1);
}
read_matrix(f1, mtx1, c1, r1);
read_matrix(f2, mtx2, c2, r2);
double* mtx3 = alloc_matrix(c1, r2);
multiplication(mtx1, mtx2, mtx3, c1, r2);
write_matrix(mtx3, c1, r2);
free(mtx1);
free(mtx2);
free(mtx3);
fclose(f1);
fclose(f2);
return 0;
}
当我用 2 个 3x3 矩阵尝试它时，前哨：
6422164 4199040（来自我设置用于检查尺寸的 2 个 printf()s）。
6422164 4199040
矩阵不适合乘法。（这无关紧要）
所以基本上它不使用3x3。
我无法弄清楚问题是什么。