On Thu, Oct 7, 2010 at 6:17 AM, Brent Yorgey <byor...@seas.upenn.edu> wrote: > The source code seems to be easy to read, but I don't think I understand > that. For me I think if I change the first line from > fib = ((map fib' [0 ..]) !!) > to > fib x = ((map fib' [0 ..]) !!) x > It should do the same thing since I think the previous version is just an > abbreviation for the second one.
Semantically, yes. And it's possible that ghc -O is clever enough to notice that. But at least under ghci's naive evaluation strategy, lambdas determine the lifetime of expressions. Any expression within a lambda will be re-evaluated each time the lambda is expanded. Thus: fib = (map fib' [0..] !!) -- fast fib = \x -> map fib' [0..] !! x -- slow fib = let memo = map fib' [0..] in \x -> memo !! x -- fast The section works because "(a %^&)" (for some operator %^&) is short for "(%^&) a" and "(%^& a)" is short for "flip (%^&) a". Sections don't expand into lambdas. In other words, in the middle expression, there is a new "map fib' [0..]" for each x, whereas in the others, it is shared between invocations. Does that make sense? Luke _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
