The code uses a variable that has not been initialized, leading to unpredictable or unintended results.
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.
Threat Mapped score: 0.0
Industry: Finiancial
Threat priority: Unclassified
CVE: CVE-2019-15900
Chain: sscanf() call is used to check if a username and group exists, but the return value of sscanf() call is not checked (CWE-252), causing an uninitialized variable to be checked (CWE-457), returning success to allow authorization bypass for executing a privileged (CWE-863).
CVE: CVE-2008-3688
Chain: A denial of service may be caused by an uninitialized variable (CWE-457) allowing an infinite loop (CWE-835) resulting from a connection to an unresponsive server.
CVE: CVE-2008-0081
Uninitialized variable leads to code execution in popular desktop application.
CVE: CVE-2007-4682
Crafted input triggers dereference of an uninitialized object pointer.
CVE: CVE-2007-3468
Crafted audio file triggers crash when an uninitialized variable is used.
CVE: CVE-2007-2728
Uninitialized random seed variable used.
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| Phase | Note |
|---|---|
| Implementation | In C, using an uninitialized char * in some string libraries will return incorrect results, as the libraries expect the null terminator to always be at the end of a string, even if the string is empty. |
Intro: This code prints a greeting using information stored in a POST request:
Body: 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, $nameArray will remain uninitialized. This will cause an error when the array is accessed to print the greeting message, which could lead to further exploit.
if (isset($_POST['names'])) { $nameArray = $_POST['names']; } echo "Hello " . $nameArray['first'];Intro: The following switch statement is intended to set the values of the variables aN and bN before they are used:
Body: 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.
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);Intro: 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.
Body: When the printf() is reached, test_string might be an unexpected address, so the printf might print junk strings (CWE-457). To fix this code, there are a couple approaches to making sure that test_string has been properly set once it reaches the printf(). One solution would be to set test_string to an acceptable default before the conditional:
char *test_string; if (i != err_val) { test_string = "Hello World!"; } printf("%s", test_string);