Hi all, does someone know why this program works

------------------------ import Concurrent

loopM = sequence_ . repeat

main = do v <- newMVar 0 forkIO (loopM (modifyMVar_ v (return . (1+)))) -- forkIO (loopM (modifyMVar_ v (return . (1+)))) loopM (readMVar v >>= print) -------------------------

but uncommenting the commented line makes the program fail with "stack overflow" ?

There's four interesting pieces of code here: M1 = the code in main before the forks M2 = the body of the loop at the end of main A = the body of the loop in the first forked thread B = the body of the loop in the second forked thread One possible execution sequence is: M1 A B A B A B A B A B M2 ... With this, the code in A and B will build a bigger and bigger suspended computation of the form 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 and then when M2 runs it will evaluate this computation and update it. This evaluation requires stack space proportional to the number of As and Bs. A more likely execution sequence would have each thread run for a significant timeslice so you'd see perhaps 10000 A's then 10000 B's then 10000 M2s. M1 A A A A A A A A A ... A A A A A A B B B B B B B B B B B B B B B B B B ... B B B B B B B B M2 M2 M2 M2 M2 M2 M2 M2 M2 M2 That first M2 will perform the same evaluation as before (i.e., adding all those ones together) but, in this more likely scenario, will require a very, very large stack. I conjecture that the problem would go away if you added appropriate uses of seq or $! to your program so that the value stored in the MVar was always fully evaluated. I leave writing a function which does this as an exercise - though not a completely trivial one since both monads and higher order functions make blindly adding $! everywhere ineffective. -- Alastair Reid alastair@reid-consulting-uk.ltd.uk Reid Consulting (UK) Limited http://www.reid-consulting-uk.ltd.uk/alastair/