#include <stdio.h>
#include <stdlib.h>
void multiplyMatrix (int **first, int **second, int **multiply);
int m, n, p, q, i, c, d, k, sum = 0;
int main()
{
int **first, **second, **multiply;
printf("Enter the number of rows and columns of first matrix\n");
scanf("%d%d", &m, &n);
first = (int **) malloc(m * sizeof(int *));
for(i = 0 ; i < n ; i++){
first[i]=(int *)malloc(m * sizeof(int *));
}
printf("Enter the elements of first matrix\n");
for (c = 0; c < m; c++)
for (d = 0; d < n; d++)
scanf("%d", &first[c][d]);
printf("Enter the number of rows and columns of second matrix\n");
scanf("%d%d", &p, &q);
second = (int **) malloc(p * sizeof(int *));
for(i = 0 ; i < q ; i++){
second[i]=(int *) malloc(p * sizeof(int *));
}
if (n != p)
printf("Matrices with entered orders can't be multiplied with each other.\n");
else
{
printf("Enter the elements of second matrix\n");
for (c = 0; c < p; c++)
for (d = 0; d < q; d++)
scanf("%d", &second[c][d]);
/*for (c = 0; c < m; c++) {
for (d = 0; d < q; d++) {
for (k = 0; k < p; k++) {
sum = sum + first[c][k]*second[k][d];
}
multiply[c][d] = sum;
sum = 0;
}
}*/
multiplyMatrix(first, second, multiply);
printf("Product of entered matrices:-\n");
for (c = 0; c < m; c++) {
for (d = 0; d < q; d++)
printf("%d\t", multiply[c][d]);
printf("\n");
}
}
return 0;
}
void multiplyMatrix (int **first, int **second, int **multiply)
{
for (c = 0; c < m; c++) {
for (d = 0; d < q; d++) {
for (k = 0; k < p; k++) {
sum = sum + first[c][k]*second[k][d];
}
multiply[c][d] = sum;
sum = 0;
}
}
}
我想写的程序应该是这样的:程序要求用户输入2个矩阵的大小和元素(或者你可以称之为2d数组)。然后它将乘以那些矩阵并打印出答案。
我得到的问题:我使用指针和malloc函数来动态分配矩阵。为了乘法,我创建了一个名为&#34; multiplyMatrix&#34;的函数。我得到了一个警告,其中的一个论点在decleration。这是警告:
警告:&#39;乘以&#39;可以在此功能中使用未初始化。
因此初始化此参数存在某种问题。我觉得答案很简单,但与此同时我无法找到解决方案。
答案 0 :(得分:1)
您尚未分配乘法矩阵使用的内存 - 因此它被标记为未初始化。
您还需要在分配第一个和第二个矩阵时查看如何使用行和列值,例如:
first = (int **) malloc(m * sizeof(int *));
for(i = 0 ; i < m ; i++){
first[i]=(int *)malloc(n * sizeof(int *));
}
(包含wildplasser发表的评论)
这将允许首先作为第一[row] [col]
进行访问答案 1 :(得分:0)
变量multiply
在main()中声明,但它永远不会指向任何东西。它需要以与first
和second
相同的方式创建,但它不需要具有&#39;填写的值。
答案 2 :(得分:0)
改进代码的建议:
main
中的代码。#include <stdio.h>
#include <stdlib.h>
int** createMatrix(int rows, int cols)
{
int i;
int** mat = malloc(sizeof(*mat)*rows);
for ( i = 0; i < rows; ++i )
mat[i] = malloc(sizeof(*mat[i])*cols);
return mat;
}
void readMatrix(int** mat, int rows, int cols)
{
int r;
int c;
for ( r = 0; r < rows; ++r )
for ( c = 0; c < cols; ++c )
scanf("%d", &mat[c][c]);
}
void deleteMatrix(int** mat, int rows)
{
int i;
for ( i = 0; i < rows; ++i )
free(mat[i]);
free(mat);
}
void multiplyMatrix (int **first, int **second, int **multiply,
int frows, int fcols, int scols)
{
int sum = 0;
int r;
int c;
int k;
for (r = 0; r < frows; r++) {
for (c = 0; c < scols; c++) {
sum = 0;
for (k = 0; k < fcols; k++) {
sum += first[r][k]*second[k][c];
}
multiply[r][c] = sum;
}
}
}
int main()
{
int m, n, p, q;
int r, c;
int **first, **second, **multiply;
printf("Enter the number of rows and columns of first matrix\n");
scanf("%d%d", &m, &n);
first = createMatrix(m, n);
printf("Enter the elements of first matrix\n");
readMatrix(first, m, n);
printf("Enter the number of rows and columns of second matrix\n");
scanf("%d%d", &p, &q);
if (n != p)
printf("Matrices with entered orders can't be multiplied with each other.\n");
else
{
second = createMatrix(p, q);
printf("Enter the elements of second matrix\n");
readMatrix(second, p, q);
multiply = createMatrix(m, q);
multiplyMatrix(first, second, multiply, m, n, q);
printf("Product of entered matrices:-\n");
for (r = 0; r < m; r++) {
for (c = 0; c < q; c++)
printf("%d\t", multiply[r][c]);
printf("\n");
}
deleteMatrix(multiply, m);
deleteMatrix(second, p);
}
deleteMatrix(first, m);
return 0;
}