From bbccc11bc8848926065915e6193fd4c6e33c85ef Mon Sep 17 00:00:00 2001 From: "Gustavo A. R. Silva" Date: Sun, 24 May 2020 15:52:38 -0500 Subject: [PATCH] audit: Use struct_size() helper in alloc_chunk One of the more common cases of allocation size calculations is finding the size of a structure that has a zero-sized array at the end, along with memory for some number of elements for that array. For example: struct audit_chunk { ... struct node { struct list_head list; struct audit_tree *owner; unsigned index; /* index; upper bit indicates 'will prune' */ } owners[]; }; Make use of the struct_size() helper instead of an open-coded version in order to avoid any potential type mistakes. So, replace the following form: offsetof(struct audit_chunk, owners) + count * sizeof(struct node); with: struct_size(chunk, owners, count) This code was detected with the help of Coccinelle. Signed-off-by: Gustavo A. R. Silva Signed-off-by: Paul Moore --- kernel/audit_tree.c | 4 +--- 1 file changed, 1 insertion(+), 3 deletions(-) diff --git a/kernel/audit_tree.c b/kernel/audit_tree.c index e49c912f862d..1b7a2f041793 100644 --- a/kernel/audit_tree.c +++ b/kernel/audit_tree.c @@ -188,11 +188,9 @@ static struct fsnotify_mark *alloc_mark(void) static struct audit_chunk *alloc_chunk(int count) { struct audit_chunk *chunk; - size_t size; int i; - size = offsetof(struct audit_chunk, owners) + count * sizeof(struct node); - chunk = kzalloc(size, GFP_KERNEL); + chunk = kzalloc(struct_size(chunk, owners, count), GFP_KERNEL); if (!chunk) return NULL;