Semaphore

Revision as of 07:45, 27 April 2014 by rosettacode>AndiPersti (Removed inappropriate "header" template)


Semaphore is a synchronization object proposed by Edsger Dijkstra. A semaphore is characterized by a natural number k. A task may atomically increase or decrease k. When k reaches 0 the tasks attempting to decrease it are blocked. These are released in an unspecified order when other tasks increase k, one per increment.

The natural number k works like a count of available slots for resources. When you (a task) want to use something (an object, a file, any resource) that can only be used by a limited number of tasks (usually one, but possibly more), you see if there are available slots (check the value of k). If there are slots available (k > 0), you take one (decrement k). When you're done with the resource, you free your slot up (increment k). If there were no slots available when you checked (k = 0), you wait until one becomes available.

A semaphore is considered a low-level synchronization primitive. They are exposed to deadlocking, like in the problem of dining philosophers.

See also mutex, a variant of semaphore.

Sample implementations / APIs

Ada

Here is an implementation of a semaphore based on protected objects. The implementation provides operations P (seize) and V (release), these names are usually used with semaphores. <lang ada> protected type Semaphore (K : Positive) is

  entry P;
  procedure V;

private

  Count : Natural := K;

end Mutex; </lang> The implementation of: <lang ada> protected body Semaphore is

  entry P when Count > 0 is
  begin
     Count := Count - 1;
  end P;
  procedure V is
  begin
     Count := Count + 1;
  end V;

end Semaphore; </lang> Use: <lang ada> declare

  S : Semaphore (5);

begin

  S.P;    -- Acquire the semaphore
  ...
  S.V;    -- Release it
  ...
  select
     S.P; -- Wait no longer than 0.5s
  or delay 0.5;
     raise Timed_Out;
  end select;
  ...
  S.V;    -- Release it

end; </lang> It is also possible to implement semaphore as a monitor task.