remote: make copy_ref() perform a deep copy
To ensure that copied refs can always be freed w/o causing a double-free, make copy_ref() perform a deep copy. Also have copy_ref() return NULL if asked to copy NULL to simplify things for the caller. Background: currently copy_ref() performs a shallow copy. This is fine for current callers who never free the result and/or only copy refs which contain NULL pointers. But copy_ref() is about to gain a new caller (guess_remote_head()) which copies refs where peer_ref is not NULL and the caller of guess_remote_head() will want to free the result. Signed-off-by: Jay Soffian <jaysoffian@gmail.com> Signed-off-by: Junio C Hamano <gitster@pobox.com>
Jay Soffian committed
Feb 27, 2009 at 14:10 UTC
7b3db095d53d19e08b27114d8706ff3be6693af7
1 file changed
+13
-5
remote.c
+13
-5
@@ -779,10 +779,18 @@ struct ref *alloc_ref(const char *name)
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static struct ref *copy_ref(const struct ref *ref)
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{
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- struct ref *ret = xmalloc(sizeof(struct ref) + strlen(ref->name) + 1);
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- memcpy(ret, ref, sizeof(struct ref) + strlen(ref->name) + 1);
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- ret->next = NULL;
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- return ret;
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+ struct ref *cpy;
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+ size_t len;
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+ if (!ref)
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+ return NULL;
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+ len = strlen(ref->name);
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+ cpy = xmalloc(sizeof(struct ref) + len + 1);
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+ memcpy(cpy, ref, sizeof(struct ref) + len + 1);
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+ cpy->next = NULL;
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+ cpy->symref = ref->symref ? xstrdup(ref->symref) : NULL;
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+ cpy->remote_status = ref->remote_status ? xstrdup(ref->remote_status) : NULL;
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+ cpy->peer_ref = copy_ref(ref->peer_ref);
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+ return cpy;
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}
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struct ref *copy_ref_list(const struct ref *ref)
@@ -801,6 +809,7 @@ static void free_ref(struct ref *ref)
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{
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if (!ref)
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return;
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+ free_ref(ref->peer_ref);
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free(ref->remote_status);
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free(ref->symref);
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free(ref);
@@ -811,7 +820,6 @@ void free_refs(struct ref *ref)
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struct ref *next;
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while (ref) {
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next = ref->next;
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- free(ref->peer_ref);
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free_ref(ref);
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ref = next;
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}