Does the following pseudocode have mutual exclusion, indefinite postponement, neither or both?

Question

For the following, I'd like to know what would happen in the case below:

//global declarations
Boolean in1=false, in2=false;
Process P1::
while(1){
  //entry protocol
  while(in2){
    in1=true;
  }
  //--critical section --
  //exit protocol
  in1=false;
}
Process P2::
while(1){
  //entry protocol
  in2=true;
  while(in1){
    in2=true;
  }
  //--critical section --
  //exit protocol
  in1=false;
}

Just a bit stumped with this at the moment, because I think it just lacks mutual exclusion - but just wanted to make sure!

Mutual exclusion is a facility provided by the OS - so no it does not — Ed Heal, Oct 05 '14 at 03:00
It looks to me like it has nothing but data races and undefined behavior. — R.. GitHub STOP HELPING ICE, Oct 05 '14 at 03:18
To achieve mutual exclusion, use the mutex() facility, using the same mutex name for both critical sections of code. otherwise, as was mentioned by @R, the result will be data races and undefined behavior. — user3629249, Oct 05 '14 at 07:56

score 0 · Answer 1 · answered Jun 01 '15 at 10:51

The pseudocode doesn't have proper mutual exclusion. Consider P1 executing first:

Since initially in2=false, P1 gets over its entry protocol and enters the critical section.
If P2 executes then, since still in1=false, P2 gets over its entry protocol and enters the critical section, too, thus no mutual exclusion.

Regarding indefinite postponement:

After P2 executed in2=true, in2 is never again set to false.
So, P1 cannot get over its entry protocol's while(in2) loop and is postponed in(de)finitely.

Does the following pseudocode have mutual exclusion, indefinite postponement, neither or both?

1 Answers1