On 17 May 2017, at 19:14, Kevin Kleinfelter <kleinfelter.gro...@gmail.com>
wrote:
> I'm stumped by the behavior of the following code fragment. Can someone help
> me understand what's happening?
>
> This code:
> (println "Holding:" (:class holding))
> (let [t (:class holding)]
> (for [x t] (println "here" x))
> (for [x t] (println "there" x))))
>
> Produces this output:
> holding: {:fundname Mutual Fund 1, :value 123.45, :class [{:class sell-me,
> :percent 100}]}
> class: [{:class sell-me, :percent 100}]
> Holding: [{:class sell-me, :percent 100}]
> there {:class sell-me, :percent 100}
>
> Why doesn't the 'for' with "here" print anything?
for produces a lazy seq so it does nothing unless you call it.
The value of a let in Clojure with more than one subexpression after the
binding is the value of the value of the last subexpression. So in your let,
the second for is evaluated because it is called to provide the return value of
the let. The first expression isn't evaluated because it isn’t part of the
return value so it isn’t called.
> I tried wrapping the fors with a do, but the output was the same:
> (println "class:" (:class holding))
> (let [t (:class holding)]
> (do
> (for [x t] (println "here" x))
> (for [x t] (println "there" x)))))
The documentation for do
https://clojuredocs.org/clojure.core/do
explains that do evaluates the expressions in order and returns the last. Your
last expression is evaluated because it is returned. The first expression is
not returned, so it isn’t being used and isn’t evaluated.
One way of evaluating both would be to write:
(let [t (:class holding)]
(doseq [a (for [x t] (println "here" x))
b (for [x t] (println "there" x))]
[a b]))
https://clojuredocs.org/clojure.core/doseq
This has a return value of nil and prints both “here” and “there”.
You could also just use a let:
(let [t (:class holding)
a (for [x t] (println "here" x))
b (for [x t] (println "there" x))]
[a b])
This prints both “here” and “there” and has a return value of a vector with two
seqs of nils.
If you don’t want the return value, the doseq is the right way to go. If you do
want a return value use a let.
Alan
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