将二维数组传递给 c 中的函数时出错
Error when passing 2D array to function in c
我正在编写一个程序来计算矩阵乘法,但它不起作用。当我调试和检查函数 printMatrixMultiplication
中数组 a
和 b
的每个值(由用户输入)时,GDB 打印出 "cannot perform pointer math on不完整类型尝试铸造。 (我已经搜索过了,但我还是不明白。)该功能仅在 main
.
中预定义输入时有效
这是我的代码
#include <stdio.h>
void input(int m, int n, double a[m][n]);
void output(int m, int n, double a[m][n]);
void printMatrixMultiplication(int row_a, int col_a, double a[row_a][col_a], int row_b, int col_b, double b[row_b][col_b]);
int main()
{
int row_a, col_a, row_b, col_b;
// get value of matrix a
printf("row_a = ");
scanf("%d", &row_a);
printf("col_a = ");
scanf("%d", &col_a);
double a[row_a][col_a];
input(row_a, col_a, a);
// output(row_a, col_a, a);
// get value of matrix b
printf("row_b = ");
scanf("%d", &row_b);
printf("col_b = ");
scanf("%d", &col_b);
double b[row_b][col_b];
input(row_b, col_b, a);
// output(row_b, col_b, a);
printMatrixMultiplication(row_a, col_a, a, row_b, col_b, b);
//test
// double a[2][2]={1,2,3,4};
// double b[2][3]={1,2,3,4,5,6};
// printMatrixMultiplication(2,2,a,2,3,b);
return 0;
}
void input(int m, int n, double a[m][n])
{
int i, j;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
scanf("%lf", &a[i][j]);
}
}
}
void output(int m, int n, double a[m][n])
{
int i, j;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
printf("%.2f ", a[i][j]);
}
printf("\n");
}
}
void printMatrixMultiplication(int row_a, int col_a, double a[row_a][col_a], int row_b, int col_b, double b[row_b][col_b])
{
if (col_a != row_b)
{
return;
}
double res[row_a][col_b]; //this matrix store results
for (int i = 0; i < row_a; i++) //the values be stored line by line, this
{ //operation is controled by i and j loops.
for (int j = 0; j < col_b; j++) //the k loop helps calculate dot_product.
{
double dot_product = 0;
for (int k = 0; k < col_a; k++)
{
dot_product += a[i][k] * b[k][j]; //ERROR HERE
}
res[i][j] = dot_product;
}
}
output(row_a, col_b, res);
}
那么,错误从何而来,如何解决?
无关紧要,但该功能没有很好地实现,所以如果可能的话,如果有人能给我一些改进的提示,我将不胜感激。
我使用的是 GCC 版本 6.3.0。
读取矩阵 b
时你的代码中有错字。
只需替换:
input(row_b, col_b, a);
与
input(row_b, col_b, b);
我正在编写一个程序来计算矩阵乘法,但它不起作用。当我调试和检查函数 printMatrixMultiplication
中数组 a
和 b
的每个值(由用户输入)时,GDB 打印出 "cannot perform pointer math on不完整类型尝试铸造。 (我已经搜索过了,但我还是不明白。)该功能仅在 main
.
这是我的代码
#include <stdio.h>
void input(int m, int n, double a[m][n]);
void output(int m, int n, double a[m][n]);
void printMatrixMultiplication(int row_a, int col_a, double a[row_a][col_a], int row_b, int col_b, double b[row_b][col_b]);
int main()
{
int row_a, col_a, row_b, col_b;
// get value of matrix a
printf("row_a = ");
scanf("%d", &row_a);
printf("col_a = ");
scanf("%d", &col_a);
double a[row_a][col_a];
input(row_a, col_a, a);
// output(row_a, col_a, a);
// get value of matrix b
printf("row_b = ");
scanf("%d", &row_b);
printf("col_b = ");
scanf("%d", &col_b);
double b[row_b][col_b];
input(row_b, col_b, a);
// output(row_b, col_b, a);
printMatrixMultiplication(row_a, col_a, a, row_b, col_b, b);
//test
// double a[2][2]={1,2,3,4};
// double b[2][3]={1,2,3,4,5,6};
// printMatrixMultiplication(2,2,a,2,3,b);
return 0;
}
void input(int m, int n, double a[m][n])
{
int i, j;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
scanf("%lf", &a[i][j]);
}
}
}
void output(int m, int n, double a[m][n])
{
int i, j;
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
printf("%.2f ", a[i][j]);
}
printf("\n");
}
}
void printMatrixMultiplication(int row_a, int col_a, double a[row_a][col_a], int row_b, int col_b, double b[row_b][col_b])
{
if (col_a != row_b)
{
return;
}
double res[row_a][col_b]; //this matrix store results
for (int i = 0; i < row_a; i++) //the values be stored line by line, this
{ //operation is controled by i and j loops.
for (int j = 0; j < col_b; j++) //the k loop helps calculate dot_product.
{
double dot_product = 0;
for (int k = 0; k < col_a; k++)
{
dot_product += a[i][k] * b[k][j]; //ERROR HERE
}
res[i][j] = dot_product;
}
}
output(row_a, col_b, res);
}
那么,错误从何而来,如何解决?
无关紧要,但该功能没有很好地实现,所以如果可能的话,如果有人能给我一些改进的提示,我将不胜感激。
我使用的是 GCC 版本 6.3.0。
读取矩阵 b
时你的代码中有错字。
只需替换:
input(row_b, col_b, a);
与
input(row_b, col_b, b);