Hello Yang ...
> Part of the code using fork is listed below:
>
> childpid = fork();
> if (childpid >= 0)
> {
> if (childpid == 0)
> {
> for (i=0; i<3; i++)
> {
> printf("mark 1!\n");
> }
> exit(0);
> }
> }
> else
> {
> for (i=0; i<3; i++)
> printf("mark 2!\n");
> wait(&status);
> exit(0);
> }
Are you sure you put the "else" part on the correct place? If the above
program is executed, you will get "mark 1" only :) But I am sure, you
were just making small mistakes on pasting the codes..
MHD Tayser and Tyler were already suggested several good advices, so I
just add another idea. What you "refer" by competition is actually a
result of scheduler picking a runnable process with highest priority on
certain time. This is the basic rule, so even another process wants to
preempt another, it must have sufficient priority to "beat" the
currently running process. Of course, currently running process will
always be kicked out from CPU if its timeslice is up (or is voluntarily
given up using yield() ) .
So, how to simulate this? Besides you need to lengthen the "for" loop (i
think while(1) is great), try to call sleep() with random interval too.
Why random interval? It will simulate a task blocking on an
unpredictable event (the event might come fast, or slow).
Random sleep will make the O(1) scheduler award different interactivity
bonus for both your parent and child task, thus it will make them
having random dynamic priority on each quantum.
Hope it helps.
regards,
Mulyadi
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