Subject:
|
Re: RCX Programming Questions and Sensor Ideas
|
Newsgroups:
|
lugnet.robotics
|
Date:
|
Wed, 11 Nov 1998 13:07:27 GMT
|
Original-From:
|
Paul Haas <paulh@%SayNoToSpam%hamjudo.com>
|
Viewed:
|
3190 times
|
| |
|
|
On Wed, 11 Nov 1998, Ben Laurie wrote:
> It isn't guaranteed to work, either, though it is likely to in the end.
> This is because two processes can conspire to always have r31 == 2 when
> you test for r31 == 1.
Oops.
I don't know how the scheduler in the RCX works. It might execute 1
opcode from each task in order. If 2 tasks get in sync, then they will
never leave the semaphore code as written.
Here's a fixed version. I added a random backoff.
> Paul Haas wrote:
> > 2. You can use a counter, because += 1 and -= 1 are atomic operations.
> >
> > int r31;
> > int done;
> > r31 = 0;
> >
> > // stuff happens.
> > // Heres our semaphore stuff.
> > done = 0; // We're not done yet.
> >
> > while ( ! done ) {
> > while ( r31 != 0 ) {
> > } ; // busy wait.
> > r31 += 1; // This operation is atomic.
> > if ( r31 == 1 ) { // Were we the only process to touch the counter?
> > // Do critical section stuff here.
> > done = 1;
> > }
else {
sleep(Random(10));
}
> > r31 -= 1; // Remove our reservation.
> > }
--
paulh@hamjudo.com http://www.hamjudo.com
The April 97 WebSight magazine describes me as "(presumably) normal".
|
|
Message has 1 Reply:
 | | Re: RCX Programming Questions and Sensor Ideas
|
| (...) Paul, Are you sure you want to sleep before decrementing r31? As written, r31 is always >= 1 during the sleep -- which prevents any other tasks from obtaining a lock on the resource during this sleep time. In other words, once any two tasks (...) (26 years ago, 11-Nov-98, to lugnet.robotics)
|
Message is in Reply To:
17 Messages in This Thread:
  
 
- Entire Thread on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
This Message and its Replies on One Page:
- Nested:
All | Brief | Compact | Dots
Linear:
All | Brief | Compact
|
|
|
|
Active threads in Robotics
|
|
|
|
|
|
