From 6479ea4a8aaa64bda350bc5324ac351900da87ae Mon Sep 17 00:00:00 2001 From: Jeff King Date: Wed, 2 Sep 2020 03:54:39 -0400 Subject: [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 non-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(). But that means that a system which _doesn't_ free the original pointer would leak it. It's not clear if any such systems exist, and that interpretation of the standard seems unlikely (I'd expect a system that doesn't deallocate to simply return the original pointer in this case). But it's easy enough to err on the safe side, and just never pass a zero size to realloc() at all. Signed-off-by: Jeff King Signed-off-by: Junio C Hamano --- wrapper.c | 7 +++++-- 1 file changed, 5 insertions(+), 2 deletions(-) diff --git a/wrapper.c b/wrapper.c index 4ff4a9c3db..bcda41e374 100644 --- a/wrapper.c +++ b/wrapper.c @@ -117,10 +117,13 @@ void *xrealloc(void *ptr, size_t size) { void *ret; + if (!size) { + free(ptr); + return xmalloc(0); + } + memory_limit_check(size, 0); ret = realloc(ptr, size); - if (!ret && !size) - ret = realloc(ptr, 1); if (!ret) die("Out of memory, realloc failed"); return ret;