realloc fails

My first attempt at realloc isn't yielding the results I'd hoped for. Can anybody point me in the right direction? (linux gcc 4.1.2)

Code:

---

int main(int argc, char *argv[])
{
  char *smallArray[300];
  char *tmpArray;
 

  if ((tmpArray = realloc (smallArray, sizeof (smallArray) + 10 )) == NULL)
  {
                    printf ("realloc failed\n");
  }
  system("PAUSE");       
  return 0;
}

---

You don't have a pointer, you have an array. "smallArray" is an array of pointers, so you can't reallocate it. Just like you can't reallocate any other array. You can use realloc on one of the pointers in the array, but not on the whole array.


Quzah.

Well that sucks. So how do I get a dynamic array that I can resize as needed?

Search the forums for something like "multidimensional arrays" or "dynamic arrays". Basically you need a pointer to a pointer. You first allocate the number of actual pointers you want. Then you allocate for each one. I just did a post a few days ago on it. Much less colorful, sad to say.


Quzah.

Quote:

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Originally Posted by beginner.c

Well that sucks. So how do I get a dynamic array that I can resize as needed?

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First malloc() a basic amount for a pointer. Now treat that pointer as if it is your array. There you go. You can realloc() it.

You don't need to call malloc first, you can do everything with realloc.

char *array = NULL;

char *temp = realloc( array, 10 * sizeof *array ); // this is identical to malloc
if ( temp != NULL ) array = temp;

Then later on, expand it with
temp = realloc( array, 20 * sizeof *array );
if ( temp != NULL ) array = temp;

That means you can just have a loop which periodically calls realloc inside the loop, starting with a NULL pointer, rather than having to prepend an initial malloc just to get things started.

Thanks, but is that increasing the number of elements of the array? That's what I need to do.

Quote:

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but is that increasing the number of elements of the array? That's what I need to do.

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Yes it does

Code:

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char *temp = realloc( array, 10 * sizeof *array );

---

This one allocated 10 bytes and makes it as string. which is something similar to array[10]

this one expands that a bit more to

Code:

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temp = realloc( array, 20 * sizeof *array );

---

And for the same array to allocated 20 more bytes to make array[30]. Thats is how it works. relloc stores the original pointer in a locatoin that was 10bytes and then allocated 20bytes and returns a nw pointer to the new memory location. This is how it works

ssharish2005

OK I must be missing something. Are you saying that the resulting array will be the equivalent of

Code:

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char *smallArray[30];

---

As I need each element to be able to store a string rather than a single char.

A while ago I posted a pretty large example on how to read a file into an array of strings. The problem is the same and you can modify the code to your needs:

http://cboard.cprogramming.com/showp...53&postcount=6

Quote:

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Originally Posted by beginner.c

OK I must be missing something. Are you saying that the resulting array will be the equivalent of

Code:

---

char *smallArray[30];

---

As I need each element to be able to store a string rather than a single char.

---

It would be just smallArray[30] rather than pointer to a char array

ssharish2005

Thanks a lot, I'll have a look through it.

Quote:

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Originally Posted by ssharish2005

Yes it does

Code:

---

char *temp = realloc( array, 10 * sizeof *array );

---

This one allocated 10 bytes and makes it as string. which is something similar to array[10]

this one expands that a bit more to

Code:

---

temp = realloc( array, 20 * sizeof *array );

---

And for the same array to allocated 20 more bytes to make array[30]. Thats is how it works. relloc stores the original pointer in a locatoin that was 10bytes and then allocated 20bytes and returns a nw pointer to the new memory location. This is how it works

ssharish2005

---

Quote:

---

Originally Posted by beginner.c

OK I must be missing something. Are you saying that the resulting array will be the equivalent of

Code:

---

char *smallArray[30];

---

As I need each element to be able to store a string rather than a single char.

---

No, no, no!
Realoc does not add 20 bytes, It changes the allocated size to the new size
In this case it changes (if successful) the 10-bytes regeon to the 20 bytes regeon and copies the 10 bytes to the new location if needed.

When the new size is less than the original size - the regeon is shrinked to the new size...