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audit: Use struct_size() helper in alloc_chunk

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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>
Signed-off-by: Paul Moore <paul@paul-moore.com>
This commit is contained in:
Gustavo A. R. Silva 2020-05-24 15:52:38 -05:00 • committed by Paul Moore
parent b3a9e3b962
commit bbccc11bc8
1 changed files with 1 additions and 3 deletions
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4
kernel/audit_tree.c
View file

@ -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;