Sem_wait() problem

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March 11th, 2013, 02:41 AM

Gyong

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Sem_wait() problem

I am using semaphores for 3 processes, P1, P2, P3
that is I can have multiple processes runnning at the same time just that
P1 and P3 cannot occur at the same time with P2.

Basically I use 2 sema, S1 for P1 and P2; S2 for P3 and P2

but in P2 how do i check for P1 and P3??

is

sem_wait(S1);
sem_wait(S2);

valid?

(I know it's not, cause if P3 is on, my P2 will block the entrance of P1 while waiting for P3 at S2 which I don't wan, is there any other ways?)

March 11th, 2013, 03:50 AM

salem

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But if all three processes always do
sem_wait(S1);
sem_wait(S2);

What is the point of having S2 at all?

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March 11th, 2013, 04:59 AM

Gyong

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Quote:

Originally Posted by salem

But if all three processes always do
sem_wait(S1);
sem_wait(S2);

What is the point of having S2 at all?

I am sorry, perhaps my question is not clear enough.
I want to have 3 processes running P1, P2 and P3 but
P2 cannot run concurrently with P1 and P3.

So my solution is using 2 sema, S1 and S2:

In P1:
sem_wait(S1);
//code;

In P3:
sem_wait(S2);
//code;

In P2:
sem_wait(S1);
sem_wait(S2);
//code

I was asking for an alternative solution to this problem, cause
when I don't want my P2 to block the entrance of P1 when P2 is waiting for S2 as P1 and P3 can occur concurrently.

As in P2 can only occur when both P1 and P3 are not occuring

March 11th, 2013, 08:29 AM

salem

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http://linux.die.net/man/3/sem_init
The first thing to do is initialise the semaphore with the value 2.
This will permit P1 and P3 to lock and unlock at will.

P2 on the other hand must tread a bit more carefully, since it needs to lock twice.
So it could either do

Code:

sem_wait(&sem);
sem_wait(&sem);

or perhaps

Code:

do {
    int s;
    sem_wait(&sem);
    if ( (s=sem_trywait(&sem)) == -1 && errno == EAGAIN ) {  // see also sem_timedwait
        sem_post(&sem);  // didn't get it a 2nd time, back off
        continue;
    } else if ( s == 0 ) {
        // success - break out
        break;
    } else {
        // something wrong - deal with it
    }
} while ( 1 );

If P2 is low priority compared to P1/P3, then backing off if it can't get the semaphore twice in quick succession seems like a good idea.

Don't forget to post() twice, when you're done

Sem_wait() problem

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