Thread: Doubts concernign reallocation a 2-d array

Well, run the code in the debugger.

Code:

Starting program: /home/sc/Documents/a.out 
2 	3 	
3 	4 	
4 	5 	
5 	6 	

Breakpoint 1, main () at foo.c:36
36	    if ((K2[i] = (int *) realloc (K1[i],(Col+2)*(sizeof(int)))) == NULL)
(gdb) print i
$9 = 0
(gdb) print K1[i]
$10 = (int *) 0x602040
(gdb) print K2[i]
$11 = (int *) 0x602040
(gdb) # now imagine realloc returned NULL on error
(gdb) set K2[i] = 0
(gdb) # now look at K1
(gdb) print K1[i]
$12 = (int *) 0x0
(gdb) # oops!!!! bye bye the last pointer to that bit of memory.
(gdb) quit
A debugging session is active.

	Inferior 1 [process 4532] will be killed.

Quit anyway? (y or n) y

Ideally, if realloc returns NULL, you should have the old pointer to hand to do something with it.
But this code just blows it away by trying to be overly clever.

I must admit, it even took me a bit to realise the problem after only looking at the code snippet and not peeking at the answer. Writes through K2 go to the same location as writes through K1. You will be assuming these go to separate arrays.

The for loop can be replaced with this, which avoids that particular problem:

Code:

for (i=0; i<Row; i++)
{
    int *temp = realloc(K1[i], (Col+2)*(sizeof(int)));
    if (temp == NULL)
        printf("\n\n Error...");
    else
        K1[i] = temp;
}

However, the next problem is that no record is kept of whether any of these failed or not. If any have failed, then the later code will cause a buffer overrun.

Originally Posted by Michal

My question is: Is it really a mistake? As I see it, even if two pointers (in this case ‘K1’ and ‘K2’) point to the same location and the latter’s value is changed it doesn’t affect the former nor the location to which it points, therefore there is no problem using the same temp pointer a couple of times in order to assume the result of realloc functions.

A realloc() call (potentially) changes what is pointed at. There's therefore an interaction you're not accounting for.