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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 <gustavoars@kernel.org>
---
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 e49c912f862d0..1b7a2f0417936 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;

--
2.26.2
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