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fsck: free tree buffers after walking unreachable objects
After calling fsck_walk(), a tree object struct may be left in the parsed state, with the full tree contents available via tree->buffer. It's the responsibility of the caller to free these when it's done with the object to avoid having many trees allocated at once. In a regular "git fsck", we hit fsck_walk() only from fsck_obj(), which does call free_tree_buffer(). Likewise for "--connectivity-only", we see most objects via traverse_one_object(), which makes a similar call. The exception is in mark_unreachable_referents(). When using both "--connectivity-only" and "--dangling" (the latter of which is the default), we walk all of the unreachable objects, and there we forget to free. Most cases would not notice this, because they don't have a lot of unreachable objects, but you can make a pathological case like this: git clone --bare /path/to/linux.git repo.git cd repo.git rm packed-refs ;# now everything is unreachable! git fsck --connectivity-only That ends up with peak heap usage ~18GB, which is (not coincidentally) close to the size of all uncompressed trees in the repository. After this patch, the peak heap is only ~2GB. A few things to note: - it might seem like fsck_walk(), if it is parsing the trees, should be responsible for freeing them. But the situation is quite tricky. In the non-connectivity mode, after we call fsck_walk() we then proceed with fsck_object() which actually does the type-specific sanity checks on the object contents. We do pass our own separate buffer to fsck_object(), but there's a catch: our earlier call to parse_object_buffer() may have attached that buffer to the object struct! So by freeing it, we leave the rest of the code with a dangling pointer. Likewise, the call to fsck_walk() in index-pack is subtle. It attaches a buffer to the tree object that must not be freed! And so rather than calling free_tree_buffer(), it actually detaches it by setting tree->buffer to NULL. These cases would _probably_ be fixable by having fsck_walk() free the tree buffer only when it was the one who allocated it via parse_tree(). But that would still leave the callers responsible for freeing other cases, so they wouldn't be simplified. While the current semantics for fsck_walk() make it easy to accidentally leak in new callers, at least they are simple to explain, and it's not a function that's likely to get a lot of new call-sites. And in any case, it's probably sensible to fix the leak first with this simple patch, and try any more complicated refactoring separately. - a careful reader may notice that fsck_obj() also frees commit buffers, but neither the call in traverse_one_object() nor the one touched in this patch does so. And indeed, this is another problem for --connectivity-only (and accounts for most of the 2GB heap after this patch), but it's one we'll fix in a separate commit. Signed-off-by: Jeff King <[email protected]> Signed-off-by: Junio C Hamano <[email protected]>
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builtin/fsck.c

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@@ -228,6 +228,8 @@ static void mark_unreachable_referents(const struct object_id *oid)
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options.walk = mark_used;
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fsck_walk(obj, NULL, &options);
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if (obj->type == OBJ_TREE)
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free_tree_buffer((struct tree *)obj);
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}
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static int mark_loose_unreachable_referents(const struct object_id *oid,

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