Double Free

Description

When a program calls free() twice with the same argument, the program's memory management data structures become corrupted. This corruption can cause the program to crash or, in some circumstances, cause two later calls to malloc() to return the same pointer. If malloc() returns the same value twice and the program later gives the attacker control over the data that is written into this doubly-allocated memory, the program becomes vulnerable to a buffer overflow attack.

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

The following code shows a simple example of a double free vulnerability.

char* ptr = (char*)malloc (SIZE);
...
if (abrt) {
  free(ptr);
}
...
free(ptr);

Double free vulnerabilities have two common (and sometimes overlapping) causes:

Although some double free vulnerabilities are not much more complicated than this example, most are spread out across hundreds of lines of code or even different files. Programmers seem particularly susceptible to freeing global variables more than once.

Example Two

While contrived, this code should be exploitable on Linux distributions that do not ship with heap-chunk check summing turned on.

#include <stdio.h>
#include <unistd.h>
#define BUFSIZE1 512
#define BUFSIZE2 ((BUFSIZE1/2) - 8)

int main(int argc, char **argv) {
  char *buf1R1;
  char *buf2R1;
  char *buf1R2;
  buf1R1 = (char *) malloc(BUFSIZE2);
  buf2R1 = (char *) malloc(BUFSIZE2);
  free(buf1R1);
  free(buf2R1);
  buf1R2 = (char *) malloc(BUFSIZE1);
  strncpy(buf1R2, argv[1], BUFSIZE1-1);
  free(buf2R1);
  free(buf1R2);
}

See Also