Scratch this question.
I think I've figured out that the step function in some people's code is
there for aesthetic reasons only. As long as the step function contains the
same code as would otherwise go directly inside the (lazy-seq ... ) there's
no difference.
That is,
(let [step (fn [a b] stuff)]
(lazy-seq (step a b)))
is the same as
(lazy-seq stuff)
The lazy documentation on the web site had made me think there was something
special about the former, i.e. the compiler was optimizing it in some
special way.
Someone please correct me if I'm still confused.
On Sun, Nov 22, 2009 at 6:06 PM, Garth Sheldon-Coulson g...@mit.edu wrote:
Hi All,
I'll confused about the hold-onto-your-head business when building lazy
seqs using lazy-seq.
The lazier documentation on the web site doesn't really clear things up
for me, though I've read it a few times.
Under what circumstances must one use a step function, and under what
circumstances it it acceptable not to?
If this is covered in depth somewhere other than the lazier page please
feel free just to send a link.
As a point of discussion, does the following function have the potential to
cause memory problems?
Some notes: *kernel* and *options* are dynamic vars I want the lazy-seq to
close over. parse-simple-atom is a function for parsing each element
(atom, **unrelated to Clojure atoms**) of the coll. type is an arg that is
used in the parsing of each element of the coll.
(defn parse-expr-coll-to-lazy-seq [coll type]
(let [kernel *kernel*
options *options*]
(lazy-seq
(binding [*kernel* kernel
*options* options]
(when-let [s (seq coll)]
(cons (parse-simple-atom (first s) type)
(parse-expr-coll-to-lazy-seq (rest s) type)))
Thanks,
Garth
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