Re: [PATCH] initramfs: Reduce hardlink hash allocation sizes

[Jan Kara]

On Tue 22-09-26 12:08:43, Thorsten Blum wrote:
> On Tue, Sep 22, 2026 at 11:45:37AM +0200, Jan Kara wrote:
> > On Sat 19-09-26 23:39:00, Thorsten Blum wrote:
> > > Each hardlink hash entry reserves N_ALIGN(PATH_MAX) bytes for its name.
> > > This makes every allocation larger than 4 KiB, placing it in the
> > > kmalloc-8k bucket even for short names.
> > > 
> > > Use a flexible array with the already validated cpio name_len to reduce
> > > allocation sizes.
> > > 
> > > Also use const for the read-only name parameter.
> > > 
> > > Signed-off-by: Thorsten Blum <blum@xxxxxxxxxx>
> > 
> > ...
> > 
> > > @@ -106,14 +106,15 @@ static char __init *find_link(int major, int minor, int ino,
> > >  			continue;
> > >  		return (*p)->name;
> > >  	}
> > > -	q = kmalloc_obj(struct hash);
> > > +
> > > +	q = kmalloc_flex(struct hash, name, nlen);
> > >  	if (!q)
> > >  		panic_show_mem("can't allocate link hash entry");
> > >  	q->major = major;
> > >  	q->minor = minor;
> > >  	q->ino = ino;
> > >  	q->mode = mode;
> > > -	strscpy(q->name, name);
> > > +	strscpy(q->name, name, nlen);
> > >  	q->next = NULL;
> > >  	*p = q;
> > >  	hardlink_seen = true;
> > 
> > What about the space for terminating \0 ? This way the stored 'name' will
> > be actually one character shorter because strscpy() will overwrite the last
> > character by \0. Or do I miss something?
> 
> do_name() guarantees that collected[name_len - 1] == '\0' before calling
> maybe_link(), so name_len already includes the NUL terminator and no
> character is overwritten.

Ah, indeed. Thanks for explanation. Feel free to add:

Reviewed-by: Jan Kara <jack@xxxxxxx>

								Honza

-- 
Jan Kara <jack@xxxxxxxx>
SUSE Labs, CR