Free of Memory not on the Heap

Description

When free() is called on an invalid pointer, the program's memory management data structures may become corrupted. This corruption can cause the program to crash or, in some circumstances, an attacker may be able to cause free() to operate on controllable memory locations to modify critical program variables or execute code.

Demonstrations

The following examples help to illustrate the nature of this weakness and describe methods or techniques which can be used to mitigate the risk.

Note that the examples here are by no means exhaustive and any given weakness may have many subtle varieties, each of which may require different detection methods or runtime controls.

Example One

In this example, an array of record_t structs, bar, is allocated automatically on the stack as a local variable and the programmer attempts to call free() on the array. The consequences will vary based on the implementation of free(), but it will not succeed in deallocating the memory.

void foo(){

  record_t bar[MAX_SIZE];

  /* do something interesting with bar */

  ...
  free(bar);

}

This example shows the array allocated globally, as part of the data segment of memory and the programmer attempts to call free() on the array.

record_t bar[MAX_SIZE]; //Global var
void foo(){


  /* do something interesting with bar */
  ...
  free(bar);

}

Instead, if the programmer wanted to dynamically manage the memory, malloc() or calloc() should have been used.

void foo(){

  record_t *bar = (record_t*)malloc(MAX_SIZE*sizeof(record_t));

  /* do something interesting with bar */

  ...
  free(bar);

}

Additionally, you can pass global variables to free() when they are pointers to dynamically allocated memory.

record_t *bar; //Global var
void foo(){

  bar = (record_t*)malloc(MAX_SIZE*sizeof(record_t));

  /* do something interesting with bar */

  ...
  free(bar);

}

See Also