Re: [Boost-users] playing around with tutorial codes to understand ioservice

Subject: Re: [Boost-users] playing around with tutorial codes to understand ioservice
From: Zachary Turner (divisortheory_at_[hidden])
Date: 2009-07-11 18:43:37

On Sat, Jul 11, 2009 at 5:35 PM, Ozgur Ozturk<ozgur7_at_[hidden]> wrote:
> Hi,
>       Thanks Zachary, yes, I have made a typo, thanks for pointing to it,
> and for the explanation.
>       But it I have the same question for your code since we have a similar
> (interleaved) output.
> What I understood was, when we call io.run, main thread is suspended until
> things assigned to it finishes.
> (that is why in the examples the final count is not printed before an
> io.run() completes, right?!)
> so how come io2.run() is invoked from main thread (before io.run() finishes)
> (and we can get interleaved results)?
> Something I must be getting wrong.
>

The typo is what's causing the interleaved results :)  io2 is never
being used.  timer2 is attached simply to io.  So you have 2 timers
both using the same io_service.  If I change the code so that timer2
uses io2, I get the following output:

[start]: t1exp: 00:00:00.984375, t2exp: 00:00:01
1: timer1, t1exp: 00:00:02.984375, t2exp: 00:00:01
2: timer1, t1exp: 00:00:01, t2exp: -00:00:01.984375
3: timer1, t1exp: 00:00:01, t2exp: -00:00:02.984375
4: timer1, t1exp: 00:00:01, t2exp: -00:00:03.984375
5: timer1, t1exp: 00:00:01, t2exp: -00:00:04.984375
Final count is 5

In this case, it will actually call print with timer2 at the very end,
but by then count is already 5, so it does not enter the if statement
and just returns immediately.