Re: 2.6.13-rc6-rt9

[Peter Zijlstra <a.p.zijlstra@chello.nl>]

On Sat, 2005-08-20 at 17:24 -0400, Jeff Dike wrote:
> On Sat, Aug 20, 2005 at 09:27:25PM +0200, Peter Zijlstra wrote:
> > Jeff, could you help us out here?
> > What exactly does uml need to get out of the calibrate delay loop?
> 
> Interrupts, it's not too demanding :-)
> 
> If it's not seeing VTALRM, then it will never leave the calibration loop.
> 
> Try stracing it and see what it's getting.

Sorry for the late reply.

Yes, that does seem to be the problem.

Even with a current -rt (2.6.14-rc2-rt5) UML does not run. The issue is
indeed (as jeff pointed out) that VTALRM is never send. The small test
programm below illustrates this.

On a non-rt kernel it completed in 1 second.
On a -rt kernel it waits at infinitum.

Kind regards,

Peter Zijlstra

---------------------------

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/time.h>
#include <signal.h>

volatile int quit = 0;

void sig_vtalrm(int signr, siginfo_t * si, void * arg)
{
        if (signr == SIGVTALRM) quit = 1;
}

int main()
{
        struct itimerval ival = {{0,0}, {1, 0}};

        struct sigaction sa;
        sa.sa_sigaction = sig_vtalrm;
        sigemptyset(&sa.sa_mask);
        sa.sa_flags = 0;
        sigaction(SIGVTALRM, &sa, NULL);

        setitimer(ITIMER_VIRTUAL, &ival, NULL);

        printf("wait\n");
        while (!quit) ;
        printf("done\n");
}


-- 
Peter Zijlstra <a.p.zijlstra@chello.nl>

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