Shashwat Black
Shashwat Black

Reputation: 1002

Addresses coincide (Pointers, C Programming)

I have two two-dimensional arrays, and i don't know why, or how, the addresses of two elements one from each array, coincide.. Here's the source code:

#include <stdio.h>

int main()
{
    int i,j,m,n,o,p,*ptr;
    printf("Enter dimension of 1st matrix: ");
    scanf("%d * %d",&m,&n);
    printf("Enter dimension of 2nd matrix: ");
    scanf("%d * %d",&o,&p);
    int *a[m][n];
    int *b[o][p];
    if (n!=o) return 0;

    printf("\nEnter 1st matrix:\n");
    for (i=0;i<m;i++)
        for (j=0;j<n;j++)
        {   printf("%d   ",(a+i*(n-1)+i+j)); scanf("%d",(a+i*(n-1)+i+j));   }

    printf("\nEnter 2nd matrix:\n");
    for (i=0;i<o;i++)
        for (j=0;j<p;j++)
        {   printf("%d   ",(b+i*(p-1)+i+j)); scanf("%d",(b+i*(p-1)+i+j));   }

    /*Printing the matrices*/
    puts("");puts("");
    for (i=0;i<m;i++)
        {for (j=0;j<n;j++)
            {   ptr = (a+i*(n-1)+i+j);
                printf(" %d ",*ptr);    }   puts("");}puts("");
    for (i=0;i<o;i++)
        {for (j=0;j<p;j++)
            {   ptr = (b+i*(p-1)+i+j);
                printf(" %d ",*ptr);    }   puts("");}
}

And here's a print screen;Screen print showing two coinciding adresses

Due to this, i have been getting errors in a simple program to calculate the product of two matrices. The question is, is this usual? Shouldn't the compiler or the OS have taken care of this?

Also, why do i have to do ptr = (a+i*(n-1)+i+j); printf(" %d ",*ptr);?
Why won't printf(" %d ",*(a+i*(n-1)+i+j)); work?

Upvotes: 2

Views: 199

Answers (4)

asaelr
asaelr

Reputation: 5456

As mention above, you probable mean to make a and b int[][] instead of int*[][]. In addition, you should not write a+i*(n-1)+i+j), but &a[i][j], or *(a+i)+j. (or another combine, like a[i]+j). the compiler should automatically translate the address to the right member of the array.

(sorry for my poor English)

p.s. Anyway, why did you write i*(n-1)+i and not simply (but, as I wrote above, wrong too) i*n?

Upvotes: 0

Sangeeth Saravanaraj
Sangeeth Saravanaraj

Reputation: 16597

I saw multiple problems so re-wrote the program as follows:

#include <stdio.h>
#include <stdlib.h>

void display(int **matrix, int r, int c)
{
    int i, j;
    for (i=0 ; i<r ; i++) {
        for (j=0 ; j<c; j++) {
            printf("%3d  ", matrix[i][j]);
        }
        printf("\n");
    }
    return;
}

int main(void)
{
    int r1, c1, r2, c2;
    int **matrix1, **matrix2;
    int i, j;

    printf("Enter r1: ");
    scanf("%d", &r1);
    printf("Enter c1: ");
    scanf("%d", &c1);

    if ((matrix1 = (int **) malloc (sizeof(int *) * r1)) == NULL) {
        printf("unable to allocate memeory \n");
        return -1;
    };
    for (i=0 ; i<r1 ; i++) {
        if ((matrix1[i] =  malloc (sizeof(int) * c1)) == NULL) {
            printf("unable to allocate memory \n");
            return -1;
        }
    }

    printf("Enter contents of matrix 1\n");
    for (i=0 ; i<r1 ; i++) {
        for (j=0 ; j<c1; j++) {
            printf("matrix1[%d][%d] :", i, j);
            scanf("%d", &matrix1[i][j]);
        }
    }

    printf("Enter r2: ");
    scanf("%d", &r2);
    printf("Enter c2: ");
    scanf("%d", &c2);

    if ((matrix2 = (int **) malloc (sizeof(int *) * r2)) == NULL) {
        printf("unable to allocate memeory \n");
        return -1;
    };
    for (i=0 ; i<r2 ; i++) {
        if ((matrix2[i] =  malloc (sizeof(int) * c2)) == NULL) {
            printf("unable to allocate memory \n");
            return -1;
        }
    }

    printf("Enter contents of matrix 2\n");
    for (i=0 ; i<r2 ; i++) {
        for (j=0 ; j<c2; j++) {
            printf("matrix1[%d][%d] :", i, j);
            scanf("%d", &matrix2[i][j]);
        }
    }

    printf("Contents of matrix 1 is as follows \n");
    display(matrix1, r1, c1);
    printf("\n\n");
    printf("Contents of matrix 2 is as follows \n");
    display(matrix2, r2, c2);

    /* now, free the contents of the matrix 1 and 2 */

    for (i=0 ; i<r1 ; i++) 
        free(matrix1[i]);
    free(matrix1);

    for (i=0 ; i<r2 ; i++) 
        free(matrix2[i]);
    free(matrix2);

    return 0;
}

Output

$ gcc 2d.c 
$ ./a.out 
Enter r1: 2
Enter c1: 2
Enter contents of matrix 1
matrix1[0][0] :1
matrix1[0][1] :2
matrix1[1][0] :3
matrix1[1][1] :4
Enter r2: 5
Enter c2: 6
Enter contents of matrix 2
matrix1[0][0] :1
matrix1[0][1] :2
matrix1[0][2] :3
matrix1[0][3] :4
matrix1[0][4] :5
matrix1[0][5] :6
matrix1[1][0] :7
matrix1[1][1] :8
matrix1[1][2] :9
matrix1[1][3] :0
matrix1[1][4] :1
matrix1[1][5] :2
matrix1[2][0] :3
matrix1[2][1] :4
matrix1[2][2] :5
matrix1[2][3] :6
matrix1[2][4] :7
matrix1[2][5] :8
matrix1[3][0] :9
matrix1[3][1] :0
matrix1[3][2] :1
matrix1[3][3] :2
matrix1[3][4] :3
matrix1[3][5] :4
matrix1[4][0] :5
matrix1[4][1] :6
matrix1[4][2] :7
matrix1[4][3] :8
matrix1[4][4] :9
matrix1[4][5] :0
Contents of matrix 1 is as follows 
  1    2  
  3    4  


Contents of matrix 2 is as follows 
  1    2    3    4    5    6  
  7    8    9    0    1    2  
  3    4    5    6    7    8  
  9    0    1    2    3    4  
  5    6    7    8    9    0  
$  

Notes:

  • when you get the rows and columns from the user, its better to use dynamic memory allocation functions like malloc() to allocate memory accordingly
  • Any malloc()'ed memory should be free()'ed
  • Your way of accessing an array cell like (a+i*(n-1)+i+j) is way too complex. When dealing with pointers/arrays, its good to maintain simplicity. Please try to stick to a[][] way of accessing an array location.

Upvotes: 1

Nuffin
Nuffin

Reputation: 3972

I think the problem is that a and b are pointers to pointers to pointers to int (int[][][]) but your code uses them as if they were pointers to pointers to int (int[][]). Because of that, even if the arrays were allocated correctly (which isn't the case) their adresses could be stored close to each other causing this unexpected behavior.

Upvotes: 0

NPE
NPE

Reputation: 500257

First of all, a and b are arrays of pointers, and the pointers are never initialized.

int *a[m][n];
int *b[o][p];

My guess is that it was meant to read:

int a[m][n];
int b[o][p];

(The rest of the code would need to be changed accordingly.)

Secondly, you're treating pointers as ints (e.g. in %d). Bear in mind that a pointer can be wider than an int. For example, on my platform pointers are 64-bit and ints are 32-bit.

Upvotes: 6

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