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load_subtree(): check earlier whether an internal node is a tree entry

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If an entry is not a tree entry, then it cannot possibly be an
internal node. But the old code checked this condition only after
allocating a leaf_node object and therefore leaked that memory.
Instead, check before even entering this branch of the code.

Signed-off-by: Michael Haggerty <mhagger@alum.mit.edu>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Michael Haggerty 2017-08-26 10:28:06 +02:00 committed by Junio C Hamano
parent 98c9897d9e
commit 4d589b87e8
1 changed files with 5 additions and 2 deletions
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7
notes.c
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@ -449,6 +449,11 @@ static void load_subtree(struct notes_tree *t, struct leaf_node *subtree,
oidcpy(&l->val_oid, entry.oid);
} else if (path_len == 2) {
/* This is potentially an internal node */
if (!S_ISDIR(entry.mode))
/* internal nodes must be trees */
goto handle_non_note;
if (get_oid_hex_segment(entry.path, 2,
object_oid.hash + prefix_len,
GIT_SHA1_RAWSZ - prefix_len) < 0)
@ -459,8 +464,6 @@ static void load_subtree(struct notes_tree *t, struct leaf_node *subtree,
xcalloc(1, sizeof(struct leaf_node));
oidcpy(&l->key_oid, &object_oid);
oidcpy(&l->val_oid, entry.oid);
if (!S_ISDIR(entry.mode))
goto handle_non_note; /* not subtree */
l->key_oid.hash[KEY_INDEX] = (unsigned char) (prefix_len + 1);
} else {
/* This can't be part of a note */