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
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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