Re: [PATCH 5/6] eventfs: get rid of dentry pointers without refcounts

Linus Torvalds
2024-01-30T22:56:24Z

On Tue, 30 Jan 2024 at 13:52, Linus Torvalds
<torvalds@linux-foundation.org> wrote:
>
> Btw, you can look at name lengths in tracefs with something stupid like this:
>
> find . | sed 's:^.*/::' | tr -c '\n' . | sort | uniq -c

Actually, since only directories have these 'ei' fields, that find
should have "-type d", and then the stats change.

Most directory filenames there are indeed longer than 8 bytes (16
bytes seems to be the most common length).

That means that whether it wins or loses in allocation size tends to
be mostly about the kmalloc sizes and the size of that structure.

And it turns out that the real memory savings there would be this patch

--- a/fs/tracefs/internal.h
+++ b/fs/tracefs/internal.h
@@ -55,15 +55,6 @@ struct eventfs_inode {
unsigned int is_events:1;
unsigned int nr_entries:30;
unsigned int ino;
- /*
- * Union - used for deletion
- * @llist: for calling dput() if needed after RCU
- * @rcu: eventfs_inode to delete in RCU
- */
- union {
- struct llist_node llist;
- struct rcu_head rcu;
- };
const char name[];
};

since with all the other cleanups, neither of those fields are actually used.

With that, the base size of 'struct eventfs_inode' actually becomes 96
bytes for me.

And at least on x86-64, the kmalloc sizes are 96 and then 128 bytes.

But that means that

- with the added name pointer, the eventfs_inode would grow to 104
bytes, and grow to that next allocation size (128)

- with the embedded name, the size is 96+strlen+1, and will also need
at least a 128 byte allocation, but any name that is shorter than 32
bytes is free

so it ends up being a wash. You're going to allocate 128 bytes
regardless. But the deallocation is simpler.

Linus