Re: proc_misc.c bug

[Andrew Morton <akpm@digeo.com>]

David Mosberger <davidm@napali.hpl.hp.com> wrote:
>
> interrupts_open() can easily try to kmalloc() more memory than
> supported by kmalloc.  E.g., with 16KB page size and NR_CPUS==64, it
> would try to allocate 147456 bytes.
> 
> The workaround below is to allocate 4KB per 8 CPUs.  Not really a
> solution, but the fundamental problem is that /proc/interrupts
> shouldn't use a fixed buffer size in the first place.  I suppose
> another solution would be to use vmalloc() instead.  It all feels like
> bandaids though.
> 
> 	--david
> 
> ===== fs/proc/proc_misc.c 1.71 vs edited =====
> --- 1.71/fs/proc/proc_misc.c	Sat Mar 22 22:14:49 2003
> +++ edited/fs/proc/proc_misc.c	Thu Apr 10 14:35:16 2003
> @@ -388,7 +388,7 @@
>  extern int show_interrupts(struct seq_file *p, void *v);
>  static int interrupts_open(struct inode *inode, struct file *file)
>  {
> -	unsigned size = PAGE_SIZE * (1 + NR_CPUS / 8);
> +	unsigned size = 4096 * (1 + NR_CPUS / 8);

urgh, consider me thwapped.

There continue to be a lot of places where we assume that pages are 4k (eg:
sizing of the free page reserves).  But few are as fatal as this one...

Thanks.