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From: Erik.Funkenbusch@p7.f55.n282.z1.fidonet.org (Erik Funkenbusch)
Newsgroups: comp.sys.amiga.advocacy
Subject: How do we change the scheduler? (Was
Message-ID: <1.279F39F8@mmug.mn.org>
Date: 24 Jan 91 02:48:59 GMT
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 SJ> From: jsd@boreal.rice.edu (Shawn Joel Dube)
 SJ> Date: 21 Jan 91 17:05:57 GMT
 SJ> Organization: Rice University
 SJ> Message-ID: <1991Jan21.170557.25934@rice.edu>
 SJ> Newsgroups: comp.sys.amiga.advocacy

 SJ> In article <1991Jan21.101849.13078@eecs.wsu.edu>,
 SJ> pcooper@eecs.wsu.edu (Phil Cooper - CS495) writes:
 >|> >|>
 >|> >|> *=Cycle
 >|> >|>
 >|> >|>   Amiga   Time --->
 >|> >|> Task A  |********        ********        ********        ********>
 >|> >|> Task B  |        ********        ********        ********        >
 >|> >|>
 >|> >|>   Mac  
 >|> >|> Task A  |*****               **     ****** *************  **  ***>
 >|> >|> Task B  |     ***************  *****      *             **  **   >
 >|> >|>
 >|> >|>  In my own opinion cooperative multitasking is not multitasking, 
simply
 >|> >|> because its not very transparent.
 >|> >
 >|> >Seriously, I think cooperative is better. Take the following for 
example:
 >|> >Two task are running.  One is waiting on a keypress (via OS subroutine)
 >|> >and the other is doing some serious number-crunching.  With
 SJ> the Amiga, valuable
 >|> >time is being spent doing nothing (waiting for a keypress). 
 >|>
 >|>     Of course, this is just wrong...  If one task is waiting on a key 
press,
 >|> no CPU time is consumed by it.  If the program is written correctly 
(without
 >|> a polling loop or some such nonsense), it merely waits.  When the key is
 >|> pressed, the task is signalled and it continues.  So, as you
 SJ> can see, no time
 >|> is being wasted.  Also, please check your facts before posting
 SJ> trash like this.
 >|> It really angers me (and others) when misinformation is spread under the
 >|> pretense of truth.

 SJ> Assuming the chart above is correct, time
 SJ> is allocated to each process equally.
 SJ> Any idiot can see that Task A gets just
 SJ> as much time as Task B, even if Task A
 SJ> is simply waiting for a keypress. 

 SJ> I didn't make the chart, so if it's
 SJ> wrong, don't blame me for spreading false
 SJ> information.  Blame the person who made the chart.

No, that chart is NOT quite correct, whenever a task needs a resource such 
as a key pressed or disk I/O it gives up it's processor time with a Wait() 
instructionn. in this case it's like co-operative multi-tasking, however the 
big difference is that when task A (or B) has used up it's aloted time it 
switches to the next task, wheather the program is ready or not.

 >|>
 >|>     Have a nice day...
 >|>     Phil Cooper

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