tpeng <pengt...@gmail.com> writes:
>> (defn foldr [f coll]
>> (reduce #(f %2 %1) (reverse coll)))
> but this foldr can't handle the infinite list, am i right?
Correct. In a lazily evaluated language like Haskell you can have a
combining function which doesn't always evaluate its arguments and thus
only partially uses the list. Clojure is eagerly evaluated, so we can't
do it quite the same way.
We can of course achieve much the same effect by explicitly delaying
evaluation using closures. This is how lazy-seq works. For example, if
we make the second argument to the combining function a closure:
(defn lazy-foldr [f val coll]
(if-let [[x & xs] coll]
(f x #(lazy-foldr f val xs))
val))
;; Finite seq of trues
(lazy-foldr #(and %1 (%2)) true [true true])
;; => true
;; Infinite seq of falses
(lazy-foldr #(and %1 (%2)) true (repeat false))
;; => false
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