On Sat, 25 Nov 2006 04:48:15 +0000 (GMT)
Chris Caputo <[EMAIL PROTECTED]> wrote:
>
> This patch cleans up the output of show_state/task() (aka magic-sysrq-t)
> so that free stack space is printed as appropriate based on
> CONFIG_DEBUG_STACK_USAGE.
>
> Also, without this patch the header is not aligned with the data and is
> thus confusing. Free stack is labeled as pid, pid is labeled as father,
> and so on.
>
> Signed-off-by: Chris Caputo <[EMAIL PROTECTED]>
> ---
>
> diff -uprN a/kernel/sched.c b/kernel/sched.c
> --- a/kernel/sched.c 2006-11-25 04:11:12.000000000 +0000
> +++ b/kernel/sched.c 2006-11-25 04:13:07.000000000 +0000
> @@ -4757,7 +4757,6 @@ static const char stat_nam[] = "RSDTtZX"
> static void show_task(struct task_struct *p)
> {
> struct task_struct *relative;
> - unsigned long free = 0;
> unsigned state;
>
> state = p->state ? __ffs(p->state) + 1 : 0;
> @@ -4779,10 +4778,10 @@ static void show_task(struct task_struct
> unsigned long *n = end_of_stack(p);
> while (!*n)
> n++;
> - free = (unsigned long)n - (unsigned long)end_of_stack(p);
> + printk("%5lu ", (unsigned long)n - (unsigned
> long)end_of_stack(p));
> }
> #endif
> - printk("%5lu %5d %6d ", free, p->pid, p->parent->pid);
> + printk("%5d %6d ", p->pid, p->parent->pid);
This will cause the output format to be dependent upon the setting of
CONFIG_DEBUG_STACK_USAGE. So any code which attempts to parse the output
of this function will somehow need to work out whether or not the `free'
field is present.
Which is why we still print out a zero if CONFIG_DEBUG_STACK_USAGE=n.
-
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