[Haskell-cafe] How is laziness defined?

TJ tjay.dreaming at gmail.com
Mon Feb 5 00:37:36 EST 2007

On 2/5/07, ajb at spamcop.net <ajb at spamcop.net> wrote:
> Quoting TJ <tjay.dreaming at gmail.com>:
> > I would think that with 100% laziness, nothing would happen until the
> > Haskell program needed to output data to, e.g. the console. Quite
> > obviously that's not it. So how is laziness defined in Haskell?
> It means that the program behaves as if "things" are evaluated if and
> only if they are needed.  "Needed" in the Haskell sense, means "needed
> to do I/O".

So it's just IO which makes things run huh? OK that's basically what I
said there. Cool.

> > This is one of the things that just boggles my mind everytime I try to
> > wrap it around this thing called Haskell ;)
> The cool part is that for the most part, it doesn't matter.  It just
> works.  If you ever come across a case where it doesn't just work (e.g.
> if you need a tilde pattern), you'll be ready for it.

I despise using what I don't understand. It's a big problem but one
which is more insurmountable than understanding Haskell, I think...

On 2/5/07, Andrew Wagner <wagner.andrew at gmail.com> wrote:
> I found it useful to work through an example where lazy evaluation was
> important, and wrote it up in a tutorial. It may or may not help you,
> no guarantees, but here it is:
> http://www.haskell.org/haskellwiki/Haskell/Lazy_Evaluation

With the code from your tutorial,

magic :: Int -> Int -> [Int]
magic 0 _ = []
magic m n = m : (magic n (m+n))

getIt :: [Int] -> Int -> Int
getIt [] _ = 0
getIt (x:xs) 1 = x
getIt (x:xs) n = getIt xs (n-1)

and the example expression,

getIt (magic 1 1) 3

It only gets run  (starts pattern matching and all) if I do a print on
it, or run it from GHCi (which will do the"print" for me), right?



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