I'm nitpicking but, On Thu, Feb 28, 2008 at 11:44 AM, Felipe Lessa <[EMAIL PROTECTED]> wrote: > Bas van Dijk's 'always' (also called 'forever'[1])
forever a = a >> forever a always f z = f z >>= always f Forever doesn't pass the result of the action to its recursive call, always does. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
