Missing Synchronization

Description

If access to a shared resource is not synchronized, then the resource may not be in a state that is expected by the product. This might lead to unexpected or insecure behaviors, especially if an attacker can influence the shared resource.

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 intends to fork a process, then have both the parent and child processes print a single line.

static void print (char * string) {

  char * word;
  int counter;
  for (word = string; counter = *word++; ) {

    putc(counter, stdout);
    fflush(stdout);
    /* Make timing window a little larger... */

    sleep(1);

  }

}

int main(void) {

  pid_t pid;

  pid = fork();
  if (pid == -1) {
    exit(-2);
  }
  else if (pid == 0) {
    print("child\n");
  }
  else {
    print("PARENT\n");
  }
  exit(0);

}

One might expect the code to print out something like:

However, because the parent and child are executing concurrently, and stdout is flushed each time a character is printed, the output might be mixed together, such as:

See Also