On Wed, 31 Dec 2003, Ken Shan wrote: > Don't you need a (s2 -> s1) function as well, to translate the final > state back into StateT s1?
Yes, you're right: the thing actually running the stateful computation presumably expects to start it with a state of type s1 and to be able to extract from it a state of type s1 when it's done. -- Mark _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
