That example doesn't particularly "tie the knot", unless you count the fact that "break" is itself a recursive function. Usually "tie the knot" refers to some kind of circular programming, i.e. a self-referential data structure, or explicit use of Data.Function.fix to produce a recursive function (as is useful in e.g. dynamic programming) Also, you aren't understanding the laziness of the return product; instead you are still thinking of this example in terms of eager evaluation as almost every other programming language uses. A better approximation of what is going on could be represented textually as: -- break "hi\nbye" -- ( 'h' : ys, zs ) where ( ys, zs ) = break "i\nbye" -- ( 'h' : 'i' : ys, zs ) where ( ys, zs ) = break "\nbye" -- ( 'h' : 'i' : [] , "bye") Of course, if you want to get more pendantic, I've glossed over steps involving the conditional and something resembling beta-reduction. Incidentally, it's the latter part I omitted which, naively implemented, creates the space leak referred to in the original post. Best, Leon On Mon, Apr 5, 2010 at 5:19 PM, aditya siram wrote:

Hi all, For the past couple of weeks I've been trying to understand tying-the-knot style programming in Haskell.

A couple of days ago, Heinrich Apfelmus posted the following 'break' function as part of an unrelated thread: break []     = ([],[]) break (x:xs) = if x == '\n' then ([],xs) else (x:ys,zs)               where (ys,zs) = Main.break xs

I've stepped through the code manually to see if I understand and I'm hoping someone will tell me if I'm on the right track: -- break "hi\nbye" = --    let f1 = (break "i\nbye") --           = let f2 = (break "\nbye") --                    = ([],"bye") --             ('i' : fst f2, snd f2) => ('i' : [], "bye") --    ('h' : fst f1, snd f1) => ('h' : 'i' : [], "bye") --                           => ("hi","bye")

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