To reallocate an array ?

[erktek]

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

int main(int argc, char* argv[])
{
int i;
int a[1];
int *p = NULL;

p = a;

p = (int *) realloc(p,3*sizeof(int)); // here the code crashes ?

p[0] = 1;
p[1] = 2;
p[2] = 3;

for (i=0; i < 3;i++)
printf("%d\n",p[i]);

free(p);

return 0;
}

The code above crashes ? What is the reason ?
Isn't it possible to reallocate the array "a" again via pointer p ?

[Richard Bos]

er****@gmail.com wrote:

The code above crashes ? What is the reason ?
Isn't it possible to reallocate the array "a" again via pointer p ?

In a word: no. realloc() (which, btw, does not take a cast in
well-written C code) only works on null pointers or previously allocated
pointers, not on arrays, pointers to normal objects, or pointers which
have had arithmetic done on them.

Richard

[Richard Heathfield]

er****@gmail.com said:

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

int main(int argc, char* argv[])
{
int i;
int a[1];
int *p = NULL;

p = a;

p = (int *) realloc(p,3*sizeof(int));

The Standard (C89 cite) says:

void *realloc(void *ptr, size_t size);

[...] If ptr is a null pointer, the realloc function behaves like the
malloc function for the specified size. Otherwise, if ptr does not match
a pointer earlier returned by the calloc , malloc , or realloc function,
or if the space has been deallocated by a call to the free or realloc
function, the behavior is undefined."

Since you passed to realloc a pointer that was neither NULL nor a pointer
earlier returned by calloc, malloc, or realloc, the behaviour is
undefined.

Isn't it possible to reallocate the array "a" again via pointer p ?

No. Start with p = malloc(n * sizeof *p) or p = realloc(NULL, n * sizeof
*p). Then you'll be fine. Don't forget that malloc and realloc can fail,
in which case they return NULL.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Richard Heathfield]

Richard Bos said:

<snip>

realloc() [doesn't work on] pointers which
have had arithmetic done on them.

Not strictly true. I know what you mean, but you haven't said (sufficiently
precisely) what you mean. Consider:

p = malloc(n * sizeof *p);
failif(p == NULL); /* <--- just shorthand, okay? */
while(i++ < n)
{
*p++ = foo();
}
p -= n;
q = realloc(p, newcount * sizeof *p);

Here, p has most certainly had arithmetic done on it, but it can still be
passed to realloc. But if we erased the p -= n line, that would indeed
break the realloc call.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[WaterWalk]

On May 22, 8:54 pm, [e-mail] wrote:

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

int main(int argc, char* argv[])
{
int i;
int a[1];
int *p = NULL;

p = a;

p = (int *) realloc(p,3*sizeof(int)); // here the code crashes ?

p[0] = 1;
p[1] = 2;
p[2] = 3;

for (i=0; i < 3;i++)
printf("%d\n",p[i]);

free(p);

return 0;

}

The code above crashes ? What is the reason ?
Isn't it possible to reallocate the array "a" again via pointer p ?

realloc only works with pointers whose pointed to memory is allocated
dynamically, for example using malloc.