Thread: realloc() troubles

Hi, I seem to be having a lot of problems with realloc, I have some int arrays, which I'm storing values in, I'm initially declaring there size using malloc, later on in the program I need to add more elements, one at a time to the array, so i do

Code:

 test = realloc(test,no_of_elements*4)

When I then print out these arrays, it prints out elements as numbers like

Code:

4007400

yet I'm only dealing with single digit numbers

If I comment out the realloc statements, and just add the elements anyway ignoring the fact I'm adding in elements out of bounds, then the output is printed out fine. Obviouslt I cannot leave it like this though as I'm out of bounds and that's unsafe.

I'm checking to see if realloc returns NULL whenever it is called, but it never does.

What is likely to cause this problem?

Thanks for any advice

I think showing us a little code is the next step.

A few things I noticed that you may want to fix, however.

1) This is no real biggie, but it's usually good practice not to cast malloc.
2) All those 4's you put in there should be replaced by sizeof(int) . That one's important.

Code:

int add_element(int *test){
test = realloc(test,(x+1)*4);
test[x] = 7;
return 0;
}

Won't work at all. First, never assign back to the same variable when using realloc(). (ie, b = realloc(a, size) is ok, but not a = realloc(a, size)) If realloc were to fail, it will not touch your previously allocated memory. So if it fails, you need to old pointer around to free it. Something like:

Code:

int *test = pointer to some memory...
int *temp;
temp = realloc(test, sizeof(int) * newsize);
if(!temp)
{
  // Realloc failed! Panic, or something, but remember:
  // test is still valid, and still must be free'd sometime.
}
// Realloc is ok.
test = temp;

Next, bigger problem: Suppose that in your original function, realloc() succeeds, but has to move your data to a different location to accomated the larger size. test's value has changed, however, you do not return this value to the caller, and your program is likely to segfault somewhere down the road. (A nasty bug to find...)

If you must change an object, pass a pointer to the object:

Code:

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

#define MIN_SIZE        10U

int *append( int *parray[], size_t size, int anynum );

int main(void) {
   int *test = malloc(MIN_SIZE * sizeof *test);
   if (test) {
      int fill;
      for (fill = 0; fill < MIN_SIZE; ++fill) {
         test[fill] = fill + 1;
      }
      /*
      ** Checking appends return value to make sure that everything worked
      */
      if (append(&test, MIN_SIZE, 11) != NULL) {
         printf("Success :)\n");
      }
      else {
         printf("Failure :(\n");
      }

      free(test);
   }
   else {
     perror("malloc");
   }

   return 0;
}

int *append( int *parray[], size_t size, int anynum ) {
   int *newarray = realloc(*parray, size + 1U);
   if (newarray) {
      *parray = newarray;
      (*parray)[size] = anynum;
   }
   else {
      free(newarray);
      newarray = NULL;
   }

   return newarray;
}

Code:

   else {
      free(newarray);
      newarray = NULL;
   }

makes no harm, but has no meaning also - newarray is already NULL in the else
So this part can be freely skipped

Originally Posted by Wiretron

Also in the above example, why is append first mentioned at the top above int main()?

It is function declaration. It describes the function prototype for the compiler.
It is needed because the function body is located after the function call.
So when compiler comes to the line where the function is called it does not know about this function nothing.

To give to compiler some knowledge about the function - function declaration is provided before actual call.

I tried out your code, and added in a loop which prints out test at the end, and I'm getting the output:

Could you post your code?

@ citizen
> else
> free(newarray);
Except if realloc fails, then newarray will be NULL anyway.

> int *newarray = realloc(*parray, size + 1U);
You still need the multiplier for the size

int *newarray = realloc(*parray, (size + 1U) * sizeof *newarray );

Ok I have got citizens example working, now say I want to print out test in a seperate function, I've called it from within main using:

Code:

print(&test,fill);

which runs the function:

Code:

int *print(int *test[],int fill){
    int i = 0;
    for(i=0;i<fill;i++){
                        printf("%d ",test[i]);
                        }
                        return 0;
}

Yet this prints outs the wierd output again, can anyone see what's going wrong?

If I change the print statement to

Code:

printf("%d ",*test[i]);

then it just crashes

print(test,fill); maybe?