The product calls free() on a pointer to a memory resource that was allocated on the heap, but the pointer is not at the start of the buffer.
This can cause the product to crash, or in some cases, modify critical program variables or execute code. This weakness often occurs when the memory is allocated explicitly on the heap with one of the malloc() family functions and free() is called, but pointer arithmetic has caused the pointer to be in the interior or end of the buffer.
Threat Mapped score: 1.8
Industry: Finiancial
Threat priority: P4 - Informational (Low)
CVE: CVE-2019-11930
function "internally calls 'calloc' and returns a pointer at an index... inside the allocated buffer. This led to freeing invalid memory."
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| Phase | Note |
|---|---|
| Implementation | N/A |
Intro: In this example, the programmer dynamically allocates a buffer to hold a string and then searches for a specific character. After completing the search, the programmer attempts to release the allocated memory and return SUCCESS or FAILURE to the caller. Note: for simplification, this example uses a hard-coded "Search Me!" string and a constant string length of 20.
Body: However, if the character is not at the beginning of the string, or if it is not in the string at all, then the pointer will not be at the start of the buffer when the programmer frees it.
#define SUCCESS (1) #define FAILURE (0) int contains_char(char c){ char *str; str = (char*)malloc(20*sizeof(char)); strcpy(str, "Search Me!"); while( *str != NULL){ if( *str == c ){ /* matched char, free string and return success */ free(str); return SUCCESS; } /* didn't match yet, increment pointer and try next char */ str = str + 1; } /* we did not match the char in the string, free mem and return failure */ free(str); return FAILURE; }Intro: This code attempts to tokenize a string and place it into an array using the strsep function, which inserts a \0 byte in place of whitespace or a tab character. After finishing the loop, each string in the AP array points to a location within the input string.
Body: Since strsep is not allocating any new memory, freeing an element in the middle of the array is equivalent to free a pointer in the middle of inputstring.
char **ap, *argv[10], *inputstring; for (ap = argv; (*ap = strsep(&inputstring, " \t")) != NULL;) if (**ap != '\0') if (++ap >= &argv[10]) break; /.../ free(ap[4]);Intro: Consider the following code in the context of a parsing application to extract commands out of user data. The intent is to parse each command and add it to a queue of commands to be executed, discarding each malformed entry.
Body: While the above code attempts to free memory associated with bad commands, since the memory was all allocated in one chunk, it must all be freed together.
//hardcode input length for simplicity char* input = (char*) malloc(40*sizeof(char)); char *tok; char* sep = " \t"; get_user_input( input ); /* The following loop will parse and process each token in the input string */ tok = strtok( input, sep); while( NULL != tok ){ if( isMalformed( tok ) ){ /* ignore and discard bad data */ free( tok ); } else{ add_to_command_queue( tok ); } tok = strtok( NULL, sep)); }