The product calls free() twice on the same memory address.

Impact: Modify MemoryExecute Unauthorized Code or Commands
When a program calls free() twice with the same argument, the program's memory management data structures may become corrupted, potentially leading to the reading or modification of unexpected memory addresses. 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. Doubly freeing memory may result in a write-what-where condition, allowing an attacker to execute arbitrary code.
ccHigh