How to properly access this realloc-ed array? - c

Question

In this code below I am trying to create an array of ints that can be accessed from the main() function, however, Address-sanitizer gives me stack-buffer-overflow-error and I cannot figure out what I am doing wrong. What am I missing?
#include <stdlib.h>
void reallocFail(int **arrayOfInts) {
*arrayOfInts = (int *)malloc(sizeof(int));
for (int i = 1; i <= 10; i++) {
*arrayOfInts = (int *)realloc(*arrayOfInts, (i) * sizeof(int));
*arrayOfInts[i - 1] = i;
}
}
int main(void) {
int *arrayOfInts;
reallocFail(&arrayOfInts);
return 0;
}

Answer 1

There is just a simple typo in *arrayOfInts[i - 1] = i;. suffix operators such as [] bind stronger than prefix operators such as *. Hence you should write:
(*arrayOfInts)[i - 1] = i;
Note also that you should check for memory reallocation failure and you can initialize *arrayOfInts to NULL as realloc(NULL, size) is equivalent to malloc(size).
Here is a modified version:
#include <string.h>
#include <stdlib.h>
int reallocFail(int **pp, int n) {
int i;
*pp = NULL;
for (i = 0; i < n; i++) {
int *p = realloc(*pp, (i + 1) * sizeof(*p));
if (p == NULL)
break;
p[i] = i + 1;
*pp = p;
}
return i;
}
int main(void) {
int *arrayOfInts = NULL;
int n = reallocFail(&arrayOfInts, 10);
for (int i = 0; i < n; i++) {
printf("%d%c", arrayOfInts[i], " \n"[i == n-1]);
}
free(arrayOfInt);
return 0;
}

Answer 2

void reallocFail(int **arrayOfInts)
{
for (int i = 1; i <= 10; i++)
{
*arrayOfInts = realloc(*arrayOfInts, i * sizeof(**arrayOfInts));
arrayOfInts[0][i - 1] = i;
}
}
int main(void) {
int *arrayOfInts = NULL;
reallocFail(&arrayOfInts);
return 0;
}
But this code can be reduced to one realloc (I understand that you test if realloc works).
IT does not check if the realloc has succeeded or failed:
void reallocFail(int **arrayOfInts)
{
int *tmp;
for (int i = 1; i <= 10; i++)
{
tmp = realloc(*arrayOfInts, i * sizeof(**arrayOfInts));
if(tmp)
{
*arrayOfInts = tmp;
arrayOfInts[0][i - 1] = i;
}
else
printf("Alloction error\n");
}
}

Answer 3

For starters this memory allocation before the for loop
*arrayOfInts = (int*)malloc(sizeof(int));
is redundant. You could just write
*arrayOfInts = NULL;
Also you need to check whether memory allocation was successful.
Also this record
*arrayOfInts[i-1] = i;
is equivalent to
*( arrayOfInts[i-1] ) = i;
but you need
( *arrayOfInts )[i-1] = i;
The function can look the following way
size_t reallocFail( int **arrayOfInts, size_t n )
{
*arrayOfInts = NULL;
size_t i = 0;
if ( n != 0 )
{
int *tmp = NULL;
do
{
tmp = realloc( *arrayOfInts, ( i + 1 ) * sizeof( int ) );
if ( tmp != NULL )
{
tmp[i] = i + 1;
*arrayOfInts = tmp;
}
} while ( tmp != NULL && ++i != n );
}
return i;
}
And the function can be called for example like
int *arrayOfInts = NULL;
size_t n = reallocFail( &arrayOfInts, 10 );
for ( size_t i = 0; i != n; i++ )
{
printf( "%d ", arrayOfInts[i] );
}
putchar( '\n' );
free( arrayOfInts );
Here is a demonstration program.
#include <stdio.h>
#include <stdlib.h>
size_t reallocFail( int **arrayOfInts, size_t n )
{
*arrayOfInts = NULL;
size_t i = 0;
if ( n != 0 )
{
int *tmp = NULL;
do
{
tmp = realloc( *arrayOfInts, ( i + 1 ) * sizeof( int ) );
if ( tmp != NULL )
{
tmp[i] = i + 1;
*arrayOfInts = tmp;
}
} while ( tmp != NULL && ++i != n );
}
return i;
}
int main( void )
{
int *arrayOfInts = NULL;
size_t n = reallocFail( &arrayOfInts, 10 );
for ( size_t i = 0; i != n; i++ )
{
printf( "%d ", arrayOfInts[i] );
}
putchar( '\n' );
free( arrayOfInts );
return 0;
}
The program output is
1 2 3 4 5 6 7 8 9 10
Of course there is no great sense to reallocate the memory in the loop within the function. The function just demonstrates how to manage the function realloc.