Re: proc_misc.c bug
Andrew Morton (email@example.com)
Thu, 10 Apr 2003 15:18:57 -0700
David Mosberger <firstname.lastname@example.org> 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.
> ===== 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...
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