Hi > > > do case x of > > > [] -> return 1 > > > (y:ys) -> g y >>= \temp -> f temp > > > See the rule about always binding to the previous line of a do block. > > This case then violates that. > > I assumed that the example was equivalent to : > > do case x of > [] -> return 1 > (y:ys) -> do f (<- g y) > > Shouldn't the rule work then ?
If the do was inserted, then yes, this would work. Without it, it doesn't. Perhaps this makes a restriction to not inside case/let/lambda not that severe, since usually an additional do could be inserted. Thanks Neil _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
