Dining philosophers (forks order)

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With the help of "Monitor" I made philosopher take 2 forks at one time to eat,how can I make that philosopher takes one fork and then waits for another one?

Posted 26-Apr-11 9:56am

Tirmit

Updated 26-Apr-11 11:07am

Rick Shaub

v2

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Solution 1

See
http://rosettacode.org/wiki/Dining_philosophers#C.23[^]

Posted 26-Apr-11 10:12am

tnd2000

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Sergey Alexandrovich Kryukov 26-Apr-11 17:27pm

Not a very good article; also, it does not follow original requirements (kind of cheating). Nevertheless, the reference deserves my 4.
--SA

Sergey Alexandrovich Kryukov 26-Apr-11 17:38pm

I put some details on the problem, please see my answer.
--SA

Tirmit 27-Apr-11 2:59am

thx ,it helped me,but now i`ve got 1 exception sometimes:
cause 1 philosopher can take his fork of one fork of his two neighbour's.And from time to time the is exception in Monitor.Exit(fork[number]);

Tirmit 27-Apr-11 3:04am

and how did they do this Monitor.TryEnter(right, ref lockedR); I mean ref lockedR ,I `ve got only the possibility to write "int milisec. Timeout"

tnd2000 27-Apr-11 11:59am

Monitor.TryEnter has many overloads, one of them takes bool as a second parameter.

Tirmit 27-Apr-11 4:21am

can I check in some way if the monitor entered or not?

tnd2000 27-Apr-11 12:08pm

That's what TryEnter(fork, locked) does. When lock is false, it means another thread has locked/entered.

Maybe what you wanted is for the thread to be blocked? If so, use Monitor.Enter instead.

If you found problems with the code, you should post it in the discussion. http://rosettacode.org/wiki/Talk:Dining_philosophers

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Sergey Alexandrovich Kryukov · Accepted Answer · 2011-04-26T11:38:00

A classical problem in its pure form looks like this: each form is a thread looping a communication tasks from a thread representing a philosopher. This communication primitive can be EventWaitHandler or… a socket. First step in the loop waits for "PickUp", a second one for "PutDown".

A simplified "cheating" variant is of a form is a shared resource protected by the regular lock. A philosopher simply gets the lock of left form then the lock of right fork, than sleeps for a wile then releases the locks in the same order. With certain timing it will create a famous "circular" deadlock when each philosopher gets one fork and waits for another one forever.

Are you supposed to demonstrate this deadlock or its resolution as well. Normally, it is required to make both versions.

Unfortunately, the task could absorb some hour of my time or so, which I cannot afford; and you're supposed to solve the problem independently. It's very good for your learning to do this exercise!

Good luck,

—SA