MiddleDebuggingCommonNot answered yet
What happens if you call free() twice on the same pointer?
This code frees the same allocation along two paths. AddressSanitizer aborts the program; without it, behaviour is erratic and the heap may be corrupted.
void process(bool early) {
int* p = static_cast<int*>(std::malloc(sizeof(int)));
*p = 42;
if (early)
std::free(p);
// ... more work ...
std::free(p);
}Identify the bug and explain the cause.
Double-free is undefined behaviour: it corrupts the heap metadata stored next to allocations and can crash, silently corrupt data, or create a security vulnerability. AddressSanitizer (-fsanitize=address) detects it instantly at runtime.
- ✗Calling
deleteon a pointer that was already deleted — same UB as double-free - ✗Not zeroing the pointer after free, then checking
if (ptr)before re-freeing — the stale non-null value passes the check - ✗Sharing a raw pointer between two containers, both of which try to free it on destruction
- →What is a use-after-free vulnerability and how does it differ from a double-free?
- →How does AddressSanitizer detect double-free without hardware support?
