Jeff King[PATCH] xrealloc: do not reuse pointer freed by zero-length realloc()
This patch fixes a bug where xrealloc(ptr, 0) can double-free and corrupt the heap on some platforms (including at least glibc).
The C99 standard says of malloc (section 7.20.3):
If the size of the space requested is zero, the behavior is implementation-defined: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.
So we might get NULL back, or we might get an actual pointer (but we're not allowed to look at its contents). To simplify our code, our xmalloc() handles a NULL return by converting it into a single-byte allocation. That way callers get consistent behavior. This was done way back in 4e7a2eccc2 (?alloc: do not return NULL when asked for zero bytes, 2005-12-29).
We also gave xcalloc() and xrealloc() the same treatment. And according to C99, that is fine; the text above is in a paragraph that applies to all three. But what happens to the memory we passed to realloc() in such a case? I.e., if we do:
ret = realloc(ptr, 0);
and "ptr" is non-NULL, but we get NULL back, is "ptr" still valid? C99 doesn't cover this case specifically, but says (section 7.20.3.4):
The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size.
So "ptr" is now deallocated, and we must only look at "ret". And since "ret" is NULL, that means we have no allocated object at all. But that's not quite the whole story. It also says:
If memory for the new object cannot be allocated, the old object is not deallocated and its value is unchanged. [...] The realloc function returns a pointer to the new object (which may have the same value as a pointer to the old object), or a null pointer if the new object could not be allocated.
So if we see a NULL return with a non-zero size, we can expect that the original object _is_ still valid. But with a zero size, it's ambiguous. The NULL return might mean a failure (in which case the object is valid), or it might mean that we successfully allocated nothing, and used NULL to represent that.
The glibc manpage for realloc() explicitly says:
[...]if size is equal to zero, and ptr is not NULL, then the call is equivalent to free(ptr).
Likewise, this StackOverflow answer:
https://stackoverflow.com/a/2135302
claims that C89 gave similar guidance (but I don't have a copy to verify it). A comment on this answer:
https://stackoverflow.com/a/2022410
claims that Microsoft's CRT behaves the same.
But our current "retry with 1 byte" code passes the original pointer again. So on glibc, we effectively free() the pointer and then try to realloc() it again, which is undefined behavior.
The simplest fix here is to just pass "ret" (which we know to be NULL) to the follow-up realloc(). That does mean that a system which _doesn't_ free the original pointer would leak it. But that interpretation of the standard seems unlikely (if a system didn't deallocate in this case, I'd expect it to simply return the original pointer). If it turns out to be an issue, we can handle the "!size" case up front instead, before we call realloc() at all.
Signed-off-by: Jeff King <peff@peff.net> --- wrapper.c | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-)
Show changes to wrapper.c +1 −1
diff --git a/wrapper.c b/wrapper.c index 4ff4a9c3db..b0d375beee 100644 --- a/wrapper.c +++ b/wrapper.c @@ -120,7 +120,7 @@ void *xrealloc(void *ptr, size_t size) memory_limit_check(size, 0); ret = realloc(ptr, size); if (!ret && !size) - ret = realloc(ptr, 1); + ret = realloc(ret, 1); if (!ret) die("Out of memory, realloc failed"); return ret;
-- 2.28.0.844.g468840f815
