pointer to pointer, I am lost

Question

This is my code, it compile and runs:

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

int main() 
{ 
    int i,j; 
    int** p = (int **)malloc(2 * sizeof(int *));

    p[0] = (int *)malloc(2 * sizeof(int)); 
    p[1] = p[0];

    for(i = 0; i < 2; i++) 
    {
            for(j = 0; j < 2; j++) 
            {
                p[i][j] = i + j; 
            }
    } 

    printf("p[0][0] = %d",p[0][0]); 

    return 0; 
}

When this program is run, I got: P[0][0] = 1. Why not P[0][0] = 0?

[Never cast the result of malloc](http://stackoverflow.com/q/605845/4228131) — WedaPashi, Jun 03 '15 at 05:49

score 7 · Answer 1 · edited Jun 03 '15 at 04:05

7

You are making p[0] and p[1] point to the same memory location. Later on, you run this loop:

for(i = 0; i < 2; i++) 
            for(j = 0; j < 2; j++) 
                    p[i][j] = i + j;

which sets p[0][0] to 0. However, it is followed by setting p[1][0] to 1. Since p[0] and p[1] are equal, p[0][0] ends up being 1 finally.

What you likely intended instead of

p[0] = malloc(2 * sizeof(int)); 
p[1] = p[0];

was a separate malloc for p[1] as well. Like this:

p[0] = malloc(2 * sizeof(int)); 
p[1] = malloc(2 * sizeof(int));

edited Jun 03 '15 at 04:05

Gopi

19,784
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answered Jun 03 '15 at 04:01

Pradhan

16,391
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1

so based on your explanation, we need to allocate different memory for p[1]? p[1] = (int *)malloc(2 * sizeof(int)); – Anonymous Duck Jun 03 '15 at 04:04
@maku : Yes else they are just pointer and will point to same memory blocks. – cruxion effux Jun 03 '15 at 05:19

Sourav Kanta · Answer 2 · 2015-06-03T05:50:31.697

4

P[1] and p[0] are being assigned to the same memory location. So any change in p[1] changes p[0].You should allocate separate space for p[1] .

p[1]=malloc(2 * sizeof(int));

edited Jun 03 '15 at 05:50

answered Jun 03 '15 at 04:04

Sourav Kanta

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Oh sorry. An old habit of mine. Thank you . – Sourav Kanta Jun 03 '15 at 05:50

pointer to pointer, I am lost

2 Answers2