Hello, You could do:
(f =<< x) =<< y ? - jeremy At Thu, 12 Feb 2009 23:36:19 +0000, Edsko de Vries wrote: > > Hi, > > I can desugar > > do x' <- x > f x' > > as > > x >>= \x -> f x' > > which is clearly the same as > > x >>= f > > However, now consider > > do x' <- x > y' <- y > f x' y' > > desugared, this is > > x >>= \x -> y >>= \y' -> f x' y' > > I can simplify the second half to > > x >>= \x -> y >>= f x' > > but now we are stuck. I feel it should be possible to write something like > > x ... y ... f > > or perhaps > > f ... x ... y > > the best I could come up with was > > join $ return f `ap` x `ap` y > > which is not terrible but quite as easy as I feel this should be. Any hints? > > Edsko > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
