Segmentation fault while dynamic allocation of 2D array in C

Question

I am a newbie to C programming (relearning it after a long time) . I am trying to dynamically allocate memory to a 2D array using malloc. I have tried following the answers on stackoverflow like this and this. But I still get the segmentation fault.

My code is as below

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

void allocate2DArray(int **subset, int a, int b)
{
  subset = (int **)malloc( a * sizeof(int *));
  int i,j;

  for(i = 0 ; i < a ; i++)
    subset[i] = (int *) malloc( b * sizeof(int));

  for(i = 0 ; i < a ; i++)
     for(j = 0 ; j < b ; j++)
        subset[i][j] = 0;
}

int main()
{
  int **subset;
  int  a = 4, b = 4;
  allocate2DArray(subset, a, b);
  int i,j;
  for( i = 0 ; i < a  ; i++)
  { 
     for( j = 0 ; j < b ; j++)
     {
        printf("%d ", subset[i][j]);
     }
     printf("\n");
  }
}

When I comment the lines to print the array, it doens't give any error and program executes without segmentation fault. Please help me understand where I am going wrong.

Function arguments in C are passed by value. So your function call in main does nothing (except for leaking memory). — Kerrek SB
– Kerrek SB, Commented Jun 27, 2013 at 23:46

ctn · Accepted Answer · 2013-06-27 23:47:14Z

2

All problems in computer science can be solved by another level of indirection:

void allocate2DArray(int ***p, int a, int b)
{
    int **subset;
    *p = (int **) malloc(a * sizeof(int *));
    subset = *p;

// ...
allocate2DArray(&subset, a, b);

answered Jun 27, 2013 at 23:47

ctn

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3 Comments

quirkystack Over a year ago

Thanks @ctn ..that worked. I also tried it without using the int **subset. like this..

*p = (int **)malloc(a * sizeof(int *));  for( i = 0 ; i < a ; i++)  {    (*p)[i] = (int *)malloc(b * sizeof(int));  }  for(i = 0 ; i < a ; i++)   for(j = 0 ; j < b ; j++)     (*p)[i][j] = i+j; }

Couldn't get why this is not working. Any idea why?

ctn Over a year ago

@quirkystack For me

void allocate2DArray(int ***p, int a, int b) { int i, j; *p = (int**)malloc(a*sizeof(int *)); for(i=0; i<a; i++) (*p)[i] = (int*)malloc(b*sizeof(int)); for(i=0; i<a; i++) for(j= 0; j<b; j++) (*p)[i][j] = i+j; }

with main() { /* ... */ int **subset; allocate2DArray(&subset, a, b); /* ... */ } is working

quirkystack Over a year ago

my mistake.. didn't add paranthesis around *p in my actual program. Thanks

Emanuele Paolini · Accepted Answer · 2013-06-27 23:47:07Z

0

you must pass a int ***subset to the allocation function. This because arguments are passed by value.

answered Jun 27, 2013 at 23:47

Emanuele Paolini

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Comments

jh314 · Accepted Answer · 2013-06-27 23:48:09Z

0

You need this:

void allocate2DArray(int ***subset, int a, int b)

and this:

allocate2DArray(&subset, a, b);

answered Jun 27, 2013 at 23:48

jh314

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1 Comment

Abdo Over a year ago

This doesn't work. While compiling it gives two warnings: (1) assignment to ‘int ***’ from incompatible pointer type ‘int **’ (2) assignment to ‘int **’ from incompatible pointer type ‘int *’. And gives segmentation fault when executing.

Vicky · Accepted Answer · 2013-06-28 06:26:21Z

0

By using int **subset; it does not become 2D array. It is still 1D storage and just pointer to pointer.

2D array means each element of a pointer buffer must point to a buffer which is suggested by ctn. He has suggested ***ptr and *ptr is malloced which created 1st dimension of buffer. Now when you call allocate2DArray() again subset is allocated memory which create second dimension. Which validate my above statement - each element of pointer buffer must point to a buffer.

so now with suggested code -

 *p = (int **) malloc(a * sizeof(int *));

created an array each element of which point to buffer 'subset' which altogether create a true 2D array.

answered Jun 28, 2013 at 6:26

Vicky

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