Programming Tutorials

Using free() Function in C

By: Daniel Malcolm in C Tutorials on 2007-10-03  

When you allocate memory with either malloc() or calloc(), it is taken from the dynamic memory pool that is available to your program. This pool is sometimes called the heap, and it is finite. When your program finishes using a particular block of dynamically allocated memory, you should deallocate, or free, the memory to make it available for future use. To free memory that was allocated dynamically, use free(). Its prototype is

void free(void *ptr);

The free() function releases the memory pointed to by ptr. This memory must have been allocated with malloc(), calloc(), or realloc(). If ptr is NULL, free() does nothing. Sample program below demonstrates the free() function.

Using free() to release previously allocated dynamic memory.

1: /* Using free() to release allocated dynamic memory. */
3: #include <stdio.h>
4: #include <stdlib.h>
5: #include <string.h>
7: #define BLOCKSIZE 30000
9: main()
10: {
11:     void *ptr1, *ptr2;
13:     /* Allocate one block. */
15:     ptr1 = malloc(BLOCKSIZE);
17:     if (ptr1 != NULL)
18:         printf("\nFirst allocation of %d bytes successful.",BLOCKSIZE);
19:     else
20:     {
21:         printf("\nAttempt to allocate %d bytes failed.\n",BLOCKSIZE);
22:         exit(1);
23:     }
25:     /* Try to allocate another block. */
27:     ptr2 = malloc(BLOCKSIZE);
29:     if (ptr2 != NULL)
30:     {
31:         /* If allocation successful, print message and exit. */
33:         printf("\nSecond allocation of %d bytes successful.\n",
34:                BLOCKSIZE);
35:         exit(0);
36:     }
38:     /* If not successful, free the first block and try again.*/
40:     printf("\nSecond attempt to allocate %d bytes failed.",BLOCKSIZE);
41:     free(ptr1);
42:     printf("\nFreeing first block.");
44:     ptr2 = malloc(BLOCKSIZE);
46:     if (ptr2 != NULL)
47:         printf("\nAfter free(), allocation of %d bytes successful.\n",
48:                BLOCKSIZE);
49:     return(0);
50: }
First allocation of 30000 bytes successful.
Second allocation of 30000 bytes successful.

AANALYSIS: This program tries to dynamically allocate two blocks of memory. It uses the defined constant BLOCKSIZE to determine how much to allocate. Line 15 does the first allocation using malloc(). Lines 17 through 23 check the status of the allocation by checking to see whether the return value was equal to NULL. A message is displayed, stating the status of the allocation. If the allocation failed, the program exits. Line 27 tries to allocate a second block of memory, again checking to see whether the allocation was successful (lines 29 through 36). If the second allocation was successful, a call to exit() ends the program. If it was not successful, a message states that the attempt to allocate memory failed. The first block is then freed with free() (line 41), and a new attempt is made to allocate the second block.

You might need to modify the value of the symbolic constant BLOCKSIZE. On some systems, the value of 30000 produces the following program output:

First allocation of 30000 bytes successful.
Second attempt to allocate 30000 bytes failed.
Freeing first block.
After free(), allocation of 30000 bytes successful.

On systems with virtual memory, of course, allocation will always succeed.


DO free allocated memory when you're done with it.

DON'T assume that a call to malloc(), calloc(), or realloc() was successful. In other words, always check to see that the memory was indeed allocated.

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