Fill multiarray (matrix) in C using pointers

Question

If you want to fill a matrix in C, you can use index, like this:

#include<stdio.h>

int main(){

    char m[20][30];

    int i = 0;
    int j = 0;

    while(i <= 20){
        j=0;
        while(j <= 30){

            m[i][j] = 's';
            printf("%c", m[i][j]);
            j++;

        }
        printf("\n");
        i++;
    }
}

but how can I do this with pointers?

first of all this is getting out of bounds... `i <= 20` should be `i < 20` and `j <= 30` should be `j < 30` — chrk, Mar 01 '14 at 23:47
Depends, you mean that you want to stock pointers in your matrix or to access your matrix with pointers ? — Antoine C., Mar 02 '14 at 00:00

brokenfoot · Accepted Answer · 2014-03-02T04:09:30.007

2

In your case:

To use a pointer:

char m[20][30];
char *ptr; // Your pointer 
ptr=m; // point ptr to the location where m points to

Now, replace

m[i][j] = 's';
printf("%c", m[i][j]);

with

*((char *)ptr + (i * 30) + j) = 's';
    printf("%c", *((char *)ptr + (i * 30) + j));

Why this works?

For 1-D array:

 int a[10];

You can access it like:

a[1]; //Say the 2nd element

So, in terms of pointers, this is equivalent to:

*((int*)a+1).

For 2-D arrays:

int b[NO_OF_ROWS][NO_OF_COLUMNS];

In generalized form, assume any elment:

b[r][c]; // let r < NO_OF_ROWS, c < NO_OF_COLUMNS be any index

is equivalent to

*((int *)b + r * NO_OF_COLUMNS  + c);

edited Mar 02 '14 at 04:09

answered Mar 01 '14 at 23:49

brokenfoot

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Second element, you mean. – Alex Reynolds Mar 01 '14 at 23:55
@Geovanny Méndez Marín: I have update my answer, see if this helps – brokenfoot Mar 02 '14 at 04:05
`(int *)`? type of `m` is `char[][30]`. Isn't `(char *)` correct? – ikh Mar 02 '14 at 04:08
In addtion, the casting is unnecessary. just `*(m + i * 30 + j)` is okay. – ikh Mar 02 '14 at 04:10
Yes that is unnecessary.. just that it helps to understand better. – brokenfoot Mar 02 '14 at 04:12

score 0 · Answer 2 · answered Mar 02 '14 at 04:19

I think that the reason of using pointers instead [] operator is it's a bit faster (in theory, anyway)... In this case, I think you want this..

#include <stdio.h>

int main()
{
    char m[20][30];

    char (*ptr1)[30];
    char *ptr2;

    for (ptr1 = m; ptr1 < m + sizeof(m) / sizeof(m[0]); ptr1++)
    {
        for (ptr2 = *ptr1; ptr2 < *ptr1 + sizeof(*ptr1) / sizeof((*ptr1)[0]); ptr2++)
        {
            *ptr2 = 's';
            printf("%c", *ptr2);
        }
        printf("\n");
    }
}

This is really epic I will take it – gMendeZm Mar 04 '14 at 16:43 — gMendeZm, Mar 04 '14 at 16:43

Fill multiarray (matrix) in C using pointers

2 Answers2

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