Realloc error in C - c

Question

I've just started C read the man page but could not find the proper answer. So the code is below
void *p = malloc(10*sizeof(int));
int *q = p;
int NUMOFINT = 10;
for (int i = 0; i < NUMOFINT; i++){
printf("%i ", q[i]);
}
void *realloc(void *p, 20*sizeof(int));
for (int i = 0; i < 21; i++){
printf("%i ", q[i]);
and it is giving this error:
malloc.c: In function ‘main’:
malloc.c:31:24: error: expected declaration specifiers or ‘...’ before numeric constant
I did not quite understand that the size format was ok for malloc() but not ok for realloc. So how do I correct the error?
Edit:
so when I make it as:
void *morep = realloc(p, 20*sizeof(int));
int *q2 = morep;
for (int i = 0; i < 20; i++){
printf("%i ", q2[i]);
}
it prints out q2[11] as 135121

Answer 1

This
void *realloc(void *p, 20*sizeof(int));
is wrong. You want
p = realloc(p, 20*sizeof(int));
Incidentally, this line is also a problem (but will compile fine):
for (int i = 0; i < 21; i++){
You allocated p to 20*sizeof(int) bytes, so element 20 is past the end of the array. You want:
for (int i = 0; i < 20; i++){

Answer 2

#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <stdio.h>
int main(int argc, char **argv)
{
int *temp; /* To use with realloc, it's more secure */
void *p = malloc(10*sizeof(int));
int *q = p;
int NUMOFINT = 10;
/* dont forget to compile with -std=99 flag, if you use this kind of for loop */
for (int i = 0; i < NUMOFINT; i++){
printf("%i ", q[i]);
}
printf("\n");
/* It's more secure to use this kind of realloc */
temp = realloc (p, 20 * sizeof(int) );
if (temp == NULL)
{
fprintf(stderr,"Reallocation failed --> %s\n", strerror(errno));
free(p);
exit(EXIT_FAILURE);
}
else
{
p = temp;
}
/* Zeroing after realloc */
for (int i = NUMOFINT; i < 21; i++)
q[i] = 0;
for (int i = 0; i < 21; i++)
{
printf("%i ", q[i]);
}
printf("\n");
return 0;
}

Answer 3

realloc can tries to expand the existing memory block in heap, if its not possible it will allocate a new separate block of memory for the new size and it will copy the data from old memory block to the new memory block also.
So if realloc returns the address same as p then it just expands the old memory block. Or else we have to free the old memory block and start using the new memory block returned by realloc.
#define NUMOFINT 10
#define NUMOFINT_NEW 20
void *p = malloc(NUMOFINT * sizeof(int));
int *q = p;
void *temp = NULL;
for (int i = 0; i < NUMOFINT; i++)
{
printf("%i ", q[i]);
}
temp = realloc(p, NUMOFINT_NEW * sizeof(int));
if (temp == NULL)
{
exit(0);
}
else if (temp != p)
{
free(p);
p = temp;
}
//else temp is equal to p
for (int i = 0; i < NUMOFINT_NEW; i++)
{
printf("%i ", q[i]);
}
Here q[0] to q[9] will have the values(0, 1, 2..9) assigned, and the rest (q[10] to q[19]) will have garbage value. We have not memset the newly allocated memory to 0 also.
you can memset also before for loop,
memset((q + NUMOFINT), 0, (NUMOFINT_NEW - NUMOFINT));
Mistakes in your program are
1) compilation error because
void *realloc(void *p, 20*sizeof(int)); This statement doesn;t looks like a function call.
2) for (int i = 0; i < 21; i++) - After memory is realloced to 20 elements, you can access only upto 19th, accessing 20th may leads to crash(undefined behaviour).
3) it prints out q2[11] as 135121 - Newly reallocated memory is neither memset to 0 nor assigned any meaningful values.