Use of Uninitialized Variable

Description

In some languages such as C and C++, stack variables are not initialized by default. They generally contain junk data with the contents of stack memory before the function was invoked. An attacker can sometimes control or read these contents. In other languages or conditions, a variable that is not explicitly initialized can be given a default value that has security implications, depending on the logic of the program. The presence of an uninitialized variable can sometimes indicate a typographic error in the 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

This code prints a greeting using information stored in a POST request:

if (isset($_POST['names'])) {
  $nameArray = $_POST['names'];
}
echo "Hello " . $nameArray['first'];

This code checks if the POST array 'names' is set before assigning it to the $nameArray variable. However, if the array is not in the POST request, $nameArray will remain uninitialized. This will cause an error when the array is accessed to print the greeting message, which could lead to further exploit.

Example Two

The following switch statement is intended to set the values of the variables aN and bN before they are used:

int aN, Bn;
switch (ctl) {

  case -1:
    aN = 0;
    bN = 0;
    break;

  case 0:
    aN = i;
    bN = -i;
    break;

  case 1:
    aN = i + NEXT_SZ;
    bN = i - NEXT_SZ;
    break;

  default:
    aN = -1;
    aN = -1;
    break;


}
repaint(aN, bN);

In the default case of the switch statement, the programmer has accidentally set the value of aN twice. As a result, bN will have an undefined value. Most uninitialized variable issues result in general software reliability problems, but if attackers can intentionally trigger the use of an uninitialized variable, they might be able to launch a denial of service attack by crashing the program. Under the right circumstances, an attacker may be able to control the value of an uninitialized variable by affecting the values on the stack prior to the invocation of the function.

Example Three

This example will leave test_string in an unknown condition when i is the same value as err_val, because test_string is not initialized (CWE-456). Depending on where this code segment appears (e.g. within a function body), test_string might be random if it is stored on the heap or stack. If the variable is declared in static memory, it might be zero or NULL. Compiler optimization might contribute to the unpredictability of this address.

char *test_string;
if (i != err_val)
{

  test_string = "Hello World!";
}
printf("%s", test_string);

char *test_string = "Done at the beginning";
if (i != err_val)
{

  test_string = "Hello World!";
}
printf("%s", test_string);

Another solution is to ensure that each branch of the conditional - including the default/else branch - could ensure that test_string is set:

char *test_string;
if (i != err_val)
{

  test_string = "Hello World!";
}
else {

  test_string = "Done on the other side!";
}
printf("%s", test_string);

See Also