Multiplication of 2D arrays, using pointers to pointers - c

Question

I've been trying to implement a program in C, where i have two 2D arrays (created with pointers) and i have to multiply them (like multiplying matrices) and store the results to a third array. The dimensions of the arrays are given by the user. I've done the code but i'm getting wrong results. Is there something wrong with my formula? I wanted to do the reading and printing with functions.
For example, when i input n = 2, m = 2 and k =2. For each element of the matrix i input
A(0)(0) = 1, A(0)(1) = 2, A(1)(0) = 3, A(1)(1) = 4
and
B(0)(0) = 1,> B(0)(1) = 2, B(1)(0) = 3, B(1)(1) = 4.
The output should've been
C(0)(0) = 7, C(0)(1) = 10, C(1)(0) = 15 and C(1)(1) = 22.
Instead the output is
C(0)(0) = 7, C(0)(1) = 10, C(1)(0) = 33, C(1)(1) = 46.
I hope it's not hard to understand, i'm not allowed to post an image yet :(
#include <stdio.h>
#include <stdlib.h>
int **read(int **x, int i, int j);
int **prod(int **x, int **y, int n, int m, int k);
void print(int **x, int r, int c);
int main()
{
int n, m, k, i, j;
printf("Give n, m and k: ");
scanf("%d%d%d", &n, &m, &k);
int **A, **B, **C;
A = (int **)malloc(n*sizeof(int *));
for (i = 0; i < n; i++)
*(A+i) = (int *)malloc(k*sizeof(int));
B = (int **)malloc(k*sizeof(int *));
for (i = 0; i < k; i++)
*(B+i) = (int *)malloc(m*sizeof(int));
C = (int **)malloc(n*sizeof(int *));
for (i = 0; i < n; i++)
*(C+i) = (int *)malloc(m*sizeof(int)
for (i = 0; i < n; i++)
for (j = 0; j < k; j++)
A = read(A, i, j);
for (i = 0; i < k; i++)
for (j = 0; j < m; j++)
B = read(B, i, j);
C = prod(A, B, n, m, k);
print(C, n, m);
}
int **read(int **x, int i, int j)
{
printf("Give value to store in cell [%d][%d]: ", i, j);
scanf("%d", &x[i][j]);
return x;
}
int **prod(int **x, int **y, int n, int m, int k)
{
int i, j, l, sum;
int **res;
for (i = 0; i < n; i++)
for (j = 0; j < k; j++)
{
sum = 0;
for (l = 0; l < m; l++)
sum = sum + (x[i][l]*y[l][j]);
res[i][j] = sum;
}
return res;
}
void print(int **x, int r, int c)
{
int i, j;
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j ++)
printf("C[%d][%d]: %d\t", i, j, x[i][j]);
printf("\n");
}
}

Answer 1

first understand the matrix without pointers
Matrix multiplication without pointers
int r1,r2,c1,c2;
int i,j,sum,k;
void matrixmul()
{
printf("Getting values for matrices");
printf("\n---------------------------");
printf("\n Enter the row for Matrix A: ");
scanf("%d",&r1);
printf("\n Enter the column for Matrix A: ");
scanf("%d",&c1);
printf("\n Enter the row for Matrix B: ");
scanf("%d",&r2);
printf("\n Enter the column for Matrix A: ");
scanf("%d",&c2);
int a[r1][c1];
int b[r2][c2];
int c[r1][c2];
if (!((c1==r2)&&(r1==c2)))
{
printf("Row column miss matching so cannot do matrix multiplication");
}
else
{
printf("\n A Matrix");
printf("\n ---------");
for (i=0;i<r1;i++)
{
for(j=0;j<c1;j++)
{
printf("\n Enter %d and %d value: ",i,j);
scanf("%d",&a[i][j]);
}
}
printf("\n B Matrix");
printf("\n ---------");
for (i=0;i<r2;i++)
{
for(j=0;j<c2;j++)
{
printf("\n Enter %d and %d value: ",i,j);
scanf("%d",&b[i][j]);
}
}
}
for(i=0;i<c1;i++)
{
for(j=0;j<r2;j++)
{
for (k=0;k<r2;k++)
{
sum = sum + a[i][k]*b[k][j];
}
c[i][j] = sum;
sum = 0;
}
}
printf("\nResultant Matrix");
printf("\n----------------\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c2;j++)
{
printf("%d\t",c[i][j]);
}
printf("\n");
}
}
void main()
{
matrixmul();
}
Output
Enter the row for Matrix A: 2
Enter the column for Matrix A: 3
Enter the row for Matrix B: 3
Enter the column for Matrix B: 2
A Matrix
Enter the 0 and 0 value:1
Enter the 0 and 1 value:2
Enter the 0 and 2 value:3
Enter the 1 and 0 value:3
Enter the 1 and 1 value:2
Enter the 1 and 2 value:11
B Matrix
Enter the 0 and 0 value:4
Enter the 0 and 1 value:5
Enter the 1 and 0 value:7
Enter the 1 and 1 value:8
Enter the 1 and 0 value:9
Enter the 1 and 1 value:10
Resultant Matrix
45 51
125 141
Matrix with pointers-in this program 2D is declared using pointer variable and get the values for the 2D array and display the values
void main()
{
int r = 2, c = 3, i, j, count;
int *a[r];//while declaring the array as pointer its number of rows are specified
for (i=0; i<r; i++)
{
a[i] = (int *)malloc(c * sizeof(int));//Allocate memory space for each row for 'c' columns
}
count = 0;
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
scanf("%d",&a[i][j] );
}
}
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
printf("%d\t ", a[i][j]);
}
printf("\n");
}
}
Output
3
2
1
1
2
3
3 2 1
1 2 3
The following is the final complete matrix multiplication using pointers
void input()
{
int r1, c1,r2,c2, i, j,k,sum;
printf("\n Enter the row for Matrix A: ");
scanf("%d",&r1);
printf("\n Enter the column for Matrix A: ");
scanf("%d",&c1);
printf("\n Enter the row for Matrix B: ");
scanf("%d",&r2);
printf("\n Enter the column for Matrix A: ");
scanf("%d",&c2);
if (!((c1==r2)&&(r1==c2)))
{
printf("Row column miss matching so cannot do matrix multiplication");
}
else
{
int *a[r1];
int *b[r2];
int *c[r1];
for (i=0; i<r1; i++)
{
a[i] = (int *)malloc(c1 * sizeof(int));
}
for (i=0; i<r2; i++)
{
b[i] = (int *)malloc(c2 * sizeof(int));
}
for (i=0; i<r1; i++)
{
c[i] = (int *)malloc(c2 * sizeof(int));
}
printf("\n A Matrix");
printf("\n ----------");
for (i = 0; i < r1; i++)
{
for (j = 0; j < c1; j++)
{
printf("\n Enter the %d and %d value:",i,j);
scanf("%d",&a[i][j] );
}
}
printf("\n B Matrix");
printf("\n ----------");
for (i = 0; i < r2; i++)
{
for (j = 0; j < c2; j++)
{
printf("\n Enter the %d and %d value:",i,j);
scanf("%d",&b[i][j]);
}
}
for(i=0;i<c1;i++)
{
for(j=0;j<r2;j++)
{
for (k=0;k<r2;k++)
{
sum = sum + a[i][k]*b[k][j];
}
c[i][j] = sum;
sum = 0;
}
}
printf("\n Resultant Matrix");
printf("\n------------------");
for (i = 0; i < r1; i++)
{
for (j = 0; j < c2; j++)
{
printf("%d\t ", c[i][j]);
}
printf("\n");
}
}
}
void main()
{
input();
}
Output
Enter the row for Matrix A: 2
Enter the column for Matrix A: 3
Enter the row for Matrix B: 3
Enter the column for Matrix B: 2
A Matrix
Enter the 0 and 0 value:1
Enter the 0 and 1 value:2
Enter the 0 and 2 value:3
Enter the 1 and 0 value:3
Enter the 1 and 1 value:2
Enter the 1 and 2 value:11
B Matrix
Enter the 0 and 0 value:4
Enter the 0 and 1 value:5
Enter the 1 and 0 value:7
Enter the 1 and 1 value:8
Enter the 2 and 0 value:9
Enter the 2 and 1 value:10
Resultant Matrix
45 51
125 141
run the above program one by one understand it then convert the final program into function as per your required,Thank you