Is it not possible for a background thread to eventually signal the condition variable?
-- vijay
There is. Gambit just detected for you that the condition you got into will guaranteedly never fix and therefore it told you it's a deadlock.2014-06-11 6:13 GMT+02:00 Vijay Mathew <vijay.the.lisper@gmail.com>:Why is the following error happening in the REPL?> (define c (make-condition-variable))> (define m (make-mutex))> (mutex-unlock! m c)*** ERROR IN ##thread-deadlock-action! -- Deadlock detectedIf I do call mutex-unlock! from a thread, it does not raise this error:> (thread-start! (make-thread (lambda () (mutex-unlock! m c) (println "ok"))))#<thread #2>1> (condition-variable-broadcast! c)=> okIs there a way to invoke mutex-unlock! from the top-level without raising the deadlock error?Thanks,--Vijay
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