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From: bruce@balilly (Bruce Lilly)
Newsgroups: comp.sys.3b1
Subject: Re: more on the HFC saga
Message-ID: <1991May29.225254.27852@blilly.UUCP>
Date: 29 May 91 22:52:54 GMT
References: <1991May23.003857.8878@blilly.UUCP> <1991May28.035153.544@highspl>
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In article <1991May28.035153.544@highspl> burris@highspl (David Burris) writes:
>
>I don't understand this and I invite you to enlighten me.
>
>If we assume a steady stream of data and understand that all the
>interrupt routines must be "polled" for each interrupt, where is the
>time savings?

Assume drivers are polled in the order
driver1->driver2->driver3->driver4. A finite amount of time is required
for each driver to determine whether it is responsible for handling an
interrupt (i.e. polling the hardware). If driver4 has the interrupt
handler, the interrupt will not be serviced until after driver1, driver2,
and driver3 have determined that there are no interrupts for them. On the
other hand, if the order is driver4->driver3->driver2->driver1, then
dirver4 can handle interrupts in a more timely manner.

Note that if there are many interrupts from different devices at the same
interrupt level, some of the earlier drivers in the chain may find that
there are interrupts for them that require servicing, further delaying
drivers that are later in the chain.

Also note that if a non-maskable interrupt, or a high-priority interrupt
occurs while early drivers are checking for an interrupt at a low
priority, later drivers in the low-priority chain will also be delayed
while the high-priority interrupt is serviced.

Once an interrupt has been serviced, there is no need for other drivers to
poll hardware unless there is another device which requires service at the
same interrupt priority.

-- 
	Bruce Lilly		blilly!balilly!bruce@sonyd1.Broadcast.Sony.COM