call these functions in a
qemu thread.
2013/8/28 Max Filippov
> On Wed, Aug 28, 2013 at 12:19 PM, she roy wrote:
> > PL061 has a register GPIOIS to control if the interrupt is
> lever-triggered
> > or edge-triggered, I will try this. Thanks.
>
> This register defines which
Thanks!
2013/8/28 Peter Maydell
> On 28 August 2013 09:01, she roy wrote:
> > qemu_set_irq(gPl061->irq, 1);
> > sleep(1);
> > qemu_set_irq(gPl061->irq, 0);
>
> Never call sleep() inside a qemu device implementation:
> this will just make the whole of QEMU
PL061 has a register GPIOIS to control if the interrupt is lever-triggered
or edge-triggered, I will try this. Thanks.
2013/8/28 Max Filippov
> On Wed, Aug 28, 2013 at 12:01 PM, she roy wrote:
> > I tested qemu_irq_pulse(gPl061->irq); the guest did not generate an
> &g
-- Forwarded message --
From: she roy
Date: 2013/8/28
Subject: Re: [Qemu-devel] trigger a gpio interrupt inside qemu
To: Max Filippov
I tested qemu_irq_pulse(gPl061->irq); the guest did not generate an
interrupt. so I changed to
qemu_set_irq(gPl061->irq, 1);
s
Is there somebody can help me to trigger a gpio interrupt inside qemu? I
wrote a simple function to trigger a interrupt in pl061.c as follow:
PL061State *gPl061;
void pl061_raise_irq()
{
qemu_set_irq(gPl061->irq, 1);
}
gPl061 is assigned in function pl061_initfn:
static int pl061_initfn(SysBusD
