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From: CDCKAB%EMUVM1.BITNET@cunyvm.cuny.edu ( Karl Brendel)
Newsgroups: comp.lang.pascal
Subject: (none)
Message-ID: <25403@adm.brl.mil>
Date: 5 Jan 91 22:06:40 GMT
Sender: news@adm.brl.mil
Lines: 34

In article <1991Jan4.145541.17737@odin.diku.dk>, ballerup@diku.dk
  (Per Goetterup) wrote:

...
>        I'm having some rather weird problems with a program of mine
>which needs to get something updated on the screen no matter what
>else the program might be doing.
>        I'm using interrupt 1C (User Timer Tick) to call my procedure which
>by the way DOESN'T call DOS (common error) - it only modifies the screen
>memory.
>        The problem is not to get it to work (it does) but what
>happens next. I start it up by saving the old vector then
>re-assigning it to point at my handler (procedure). This does indeed
>work and the screen get modified as it should while my is running
>happily along. So far so good. Then something might happen because
>sometimes the computer simply stops in the middle of the program and
>lock up (refusing warm-boots), other times it runs all the way to
>
[remainder deleted]

Your description (sans code) doesn't indicate whether you are
providing for a repeat of the interrupt while you are servicing it.
I.e., if you are still inside your routine when another 1C or 08
call occurs, what happens? While I don't think this question
addresses the problem with lockups that occur after exiting your
program, it may yield clues to some of the other behavior you
describe.

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