Re: [PATCH 2/6] i386 virtualization - Remove some dead debugging code

[Zachary Amsden <zach@vmware.com>]

Chris Wright wrote:

> * zach@vmware.com (zach@vmware.com) wrote:
>  
>
> > This code is quite dead.  Release_thread is always guaranteed that the mm has
> > already been released, thus dead_task->mm will always be NULL.
> >
> > Signed-off-by: Zachary Amsden <zach@vmware.com>
> > Index: linux-2.6.13/arch/i386/kernel/process.c
> > ===================================================================
> > --- linux-2.6.13.orig/arch/i386/kernel/process.c	2005-08-15 10:46:18.000000000 -0700
> > +++ linux-2.6.13/arch/i386/kernel/process.c	2005-08-15 10:48:51.000000000 -0700
> > @@ -421,17 +421,7 @@
> >
> > void release_thread(struct task_struct *dead_task)
> > {
> > -	if (dead_task->mm) {
> > -		// temporary debugging check
> > -		if (dead_task->mm->context.size) {
> > -			printk("WARNING: dead process %8s still has LDT? <%p/%d>\n",
> > -					dead_task->comm,
> > -					dead_task->mm->context.ldt,
> > -					dead_task->mm->context.size);
> > -			BUG();
> > -		}
> > -	}
> > -
> > +	BUG_ON(dead_task->mm);
> >    
>
> This BUG_ON() has different semantics than old dead one.  Is there a
> point?  exit_mm() has already reset this to NULL, no?
>  

Yes, completely.  This BUG() could be eliminated entirely, as trivial 
inspection shows.  I can't fathom a single reason why it should still 
exist, but the presence of it in the first place made be wonder if there 
may be some erudite reason for it.  Thus I raised the BUG to a higher 
power - obviously the LDT is gone if the MM is gone.

Zach
-
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