Nice

[Richard M. Mathews]

jackv at turnkey.tcc.com (Jack F. Vogel) writes:
>boyd at necisa.ho.necisa.oz.au (Boyd Roberts) writes:
>>richard at locus.com (Richard M. Mathews) writes:
>>>...  In many (most?) versions of Unix if an interrupt
>>>makes a process runnable at a higher priority than the running process,
>>>a context switch will be forced immediately.
> 
>>No, UNIX does not have pre-emptive scheduling.  When the above event occurs
>>the kernel context switches at the next user-mode trap, system call, sleep()
>>or the once-a-second context switch (time quantum expiry).
> 
>What will happen is that the
>interrupt will cause the kernel to run in trap(), there it will do whatever
>serving needs to be done

Thanks for the defense, Jack.  To be explicit, look for the call to
aston() in wakeup() in a VAX kernel.  This will cause a trap to occur
before ANY user mode instructions are executed; that is what I meant by
"immediately".  Before that trap returns to user mode, it will notice
runrun is set and reschedule.  In other systems (such as a few different
80*86 kernels I've seen) both traps and interrupts check runrun before
returning to user mode and can switch context immediately.

Richard M. Mathews			 Freedom for Lithuania
richard at locus.com				Laisve!
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