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* RE: How?
@ 2003-05-15 14:03 Craig Hollabaugh
  0 siblings, 0 replies; 4+ messages in thread
From: Craig Hollabaugh @ 2003-05-15 14:03 UTC (permalink / raw)
  To: linuxppc-embedded


Rakesh,


Your interrupt routine can send a signal to your user space process.

Here's what I've used

void sendSignal(void)
{
        struct task_struct *p;
        if ( pid ) {

/*
        for_each_task(p) {
                if (p->pid == pid) {
                        send_sig(SIGIO,p,1);
                        break;
                }
        }
*/

/* or better yet */

        if (p = find_task_by_pid(pid)) send_sig(SIGIO,p,1);
        /* from page 83 in ulk */
        }
}

ulk is Understanding The Linux Kernel.

Craig


--
___________________________________________________________
Dr. Craig Hollabaugh                   craig@hollabaugh.com
Author of Embedded Linux

P.O. Box 1405
Ouray, CO 81427-1405        See my "kitchen sink" resume at
970 325 4810 home            www.hollabaugh.com/resume.html
970 325 0509 office


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^ permalink raw reply	[flat|nested] 4+ messages in thread
* How ?
@ 2003-05-15  5:07 Rakesh Jagota
  2003-05-15 13:31 ` Bret Indrelee
  0 siblings, 1 reply; 4+ messages in thread
From: Rakesh Jagota @ 2003-05-15  5:07 UTC (permalink / raw)
  To: linuxppc-embedded


Hi all,

 Can anyone suggest me how we can invoke the interrupt generated in the
kernel from the user application. I am getting the interrupt in the driver, I
want to update some register in the application, when the inettrupt will be
generated, Right now I am using polloing method. Continuously I am calling
read routines.

Thanks in advance,

Rakesh Jagota


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^ permalink raw reply	[flat|nested] 4+ messages in thread

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Thread overview: 4+ messages (download: mbox.gz follow: Atom feed
-- links below jump to the message on this page --
2003-05-15 14:03 How? Craig Hollabaugh
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2003-05-15  5:07 How ? Rakesh Jagota
2003-05-15 13:31 ` Bret Indrelee
2003-05-15 13:35   ` Wolfgang Denk

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