An Operating Systems Vade Mecum

Dining philosophers 129Figure 4.6 shows each philosopher as a process and each chopstick as a resource.Each philosopher executes the following program.1 procedure Philosopher(Which : 0..4);2 var3 LeftChopstick, RightChopstick : ResourceDescriptor;4 begin5 { initialize Chopstick resource variables }6 LeftChopstick.Kind := Chopstick(Which);7 RightChopstick.Kind := Chopstick((Which+1) mod 5);8 loop { think — eat loop }9 Think; { occupy some finite amount of time in thought. }10 GetResource(LeftChopstick);11 GetResource(RightChopstick);12 Eat; { for a finite amount of time }13 ReleaseResource(LeftChopstick);14 ReleaseResource(RightChopstick);15 end; { think — eat loop }16 end Philosopher;Deadlock can occur if every philosopher tries to get chopsticks at once and each gets asfar as line 11. Each has asked for a left chopstick, and the allocator might grant all theserequests. Now every philosopher is stuck at line 11, because each right chopstick issomeone else’s left chopstick; thus each is in use. Figure 4.7 shows the resource graph inthis situation. A clever allocation policy might have prevented this disaster by keepingone chopstick free, which would have been more conservative.The dining philosophers are also subject to starvation. Let’s say that philosophers2 and 4 are currently eating. Philosophers 1 and 5 are engaged in thinking great thoughtsand will not be eating for a long time. Now philosopher 3 requests both chopsticks (oneat a time or both together). Philosopher 3 is blocked, since the chopsticks are held byphilosopher 2 and philosopher 4. Figure 4.8 shows the resource graph. Later,12534Figure 4.6 Dining philosophers