On 29 Nov 2002, Alastair Reid wrote: > Consider Haskell functions a,b,c,d and C functions A,B,C,D and a call > pattern > > a -> A -> b -> B -> c -> C -> d -> D > > That is, a calls A, calls b, calls B, calls ... > > Suppose we want A,B,C,D executed by the same foreign thread. > > Each of a,b,c,d are executed by different Haskell threads (because a > new Haskell thread is spawned for each call into Haskell) so we have > multiple Haskell threads associated with a single foreign thread.
It doesn't feel right to me that a,b,c,d are executed by different Haskell threads. Why shouldn't they all be the *same* Haskell thread? 1. Reasonable exception handling across the Haskell/foreign boundary is not currently supported, but if we imagine moving in that direction, it would seem we'd want a single thread for the given example. 2. Calling from foreign code into Haskell to a bound foreign import will require some special handling to ensure that a subsequent call out to foreign code will use the same native thread. Why couldn't this special handling select the same Haskell thread instead of creating a new one? Dean _______________________________________________ Glasgow-haskell-users mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
