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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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