我有三个错误:
read
的参数1
int * (*)[10]
但参数类型为int (*)[10][10]
以下是代码:
#include <stdio.h>
#include <stdlib.h>
void read(int *(arr[10][10]), int row, int col) { //Third error here
int i, j;
for (i = 0; i < row; i++)
for (j = 0; j < col; j++)
scanf("%d", &arr[i][j]);
}
void multiply(int arr1[10][10], int row1, int col1,
int arr2[10][10], int row2, int col2,
int *prod[10][10]) { //Third error here
int i, j, k, temp;
for (i = 0; i < row1; i++)
for (j = 0; j < col2; j++) {
temp = 0;
for (k = 0; k < col1; k++)
temp += arr1[i][k] * arr2[k][j];
prod[i][j] = temp; //First error here
}
}
void display(int arr[10][10], int row, int col) {
int i, j;
for (i = 0; i < row; i++) {
for (j = 0; j <col; j++)
printf("%d\t",arr[i][j]);
printf("\n");
}
}
int main() {
int a[10][10], b[10][10], c[10][10], m, n, p, q, i, j, k;
printf("Enter the order of matrix A:");
scanf("%d %d", &m, &n);
printf("Enter the order of matrix B:");
scanf("%d %d", &p, &q);
if (n != p) {
printf("Matrix multiplication is not possible.");
exit(0);
}
printf("Enter the elements of matrix A:\n");
read(&a, m, n); //Second error here
printf("Enter the elements of matrix B:\n");
read(&b, p, q); //Second error here
multiply(a, m, n, b, p, q, &c);
printf("Matrix A is:\n");
display(a, m, n);
printf("Matrix B is:\n");
display(b, p, q);
printf("Product matrix is:\n");
display(c, m, q);
return 0;
}
答案 0 :(得分:1)
函数参数的类型int *(arr[10][10])
与int arr[10][10]
不同。以这种方式改变原型:
void read_matrix(int arr[10][10], int row, int col)
您不应将函数命名为read
,因为它是C库用于实现FILE*
流接口的Posix系统调用的名称。将其重命名为read_matrix
并以这种方式调用它:
printf("Enter the elements of matrix A:\n");
read_matrix(a,m,n); //Second error here
printf("Enter the elements of matrix B:\n");
read_matrix(b,p,q); //Second error here
矩阵乘法的原型应该改为:
void multiply(int arr1[10][10], int row1, int col1,
int arr2[10][10], int row2, int col2,
int prod[10][10])
混淆源于数组传递给C中函数的方式:它们作为指向第一个元素的指针传递。我们说当在表达式中使用时,数组会自动衰减作为指针,并作为函数参数或返回值传递。应该以这种方式调用multiply
:
multiply(a, m, n, b, p, q, c);
请注意,对于身体有多行的所有{
循环,您应该使用大括号}
和for
。您在read
和multiply
中忽略外部循环并非严格不正确,但容易出错且不推荐。
您还应该检查scanf()
的返回值并验证矩阵尺寸。
以下是改进版本:
#include <stdio.h>
#include <stdlib.h>
int matrix_read(int arr[10][10], int row, int col) {
int i, j;
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
if (scanf("%d", &arr[i][j]) != 1)
return -1;
}
}
return 0;
}
int matrix_multiply(int arr1[10][10], int row1, int col1,
int arr2[10][10], int row2, int col2,
int prod[10][10]) {
int i, j, k, temp;
if (col1 != row2)
return -1;
for (i = 0; i < row1; i++) {
for (j = 0; j < col2; j++) {
temp = 0;
for (k = 0; k < col1; k++) {
temp += arr1[i][k] * arr2[k][j];
}
prod[i][j] = temp;
}
}
return 0;
}
void matrix_display(int arr[10][10], int row, int col) {
int i, j;
for (i = 0; i < row; i++) {
for (j = 0; j <col; j++) {
printf("%d\t", arr[i][j]);
}
printf("\n");
}
}
int main(void) {
int a[10][10], b[10][10], c[10][10], m = 0, n = 0, p = 0, q = 0;
printf("Enter the order of matrix A: ");
if (scanf("%d %d", &m, &n) != 2 || m < 1 || m > 10 || n < 1 || n > 10) {
printf("Invalid matrix size: %d x %d.\n", m, n);
exit(1);
}
printf("Enter the order of matrix B: ");
if (scanf("%d %d", &p, &q) != 2 || p < 1 || p > 10 || q < 1 || q > 10) {
printf("Invalid matrix size: %d x %d.\n", p, q);
exit(1);
}
if (n != p) {
printf("Matrix multiplication is not possible.");
exit(1);
}
printf("Enter the elements of matrix A:\n");
if (matrix_read(a, m, n)) {
printf("Invalid matrix data.\n");
exit(1);
}
printf("Enter the elements of matrix B:\n");
if (matrix_read(b, p, q)) {
printf("Invalid matrix data.\n");
exit(1);
}
if (matrix_multiply(a, m, n, b, p, q, c)) {
printf("Matrix multiplication error.\n");
exit(1);
}
printf("Matrix A is:\n");
matrix_display(a, m, n);
printf("Matrix B is:\n");
matrix_display(b, p, q);
printf("Product matrix is:\n");
matrix_display(c, m, q);
return 0;
}