Clojure: when to use memoize and when to use delay/force?

Question

I've just started learning Clojure and trying to understand the difference between 2 approaches which at first sight seem very identical.

(def func0 (delay (do 
                    (println "did some work") 
                    100)))
so.core=> (force my-delay2)
did some work
100
so.core=> (force my-delay2)
100

(defn vanilla-func [] (println "did some work") 100)
(def func1 (memoize vanilla-func))
so.core=> (func1)
did some work
100
so.core=> (func1)
100

Both approaches do some sort of function memoization. What am I missing?

I've tried to find the explanation on https://clojuredocs.org/clojure.core/delay & https://clojuredocs.org/clojure.core/memoize but couldn't.

cfrick · Accepted Answer · 2020-03-19 16:15:16Z

8

delay holds one result and you have to deref to get the result. memoize is an unbound cache, that caches the result depending on the input arguments. E.g.

user=> (def myinc (memoize (fn [x] (println x) (inc x))))
#'user/myinc
user=> (myinc 1)
1
2
user=> (myinc 1)
2

In your (argument-less) example the only difference is that you can use the result directly (no deref needed)

Classic use-cases for delay are things needed later, that would block or delay startup. Or if you want to "hide" top-level defs from the compiler (e.g. they do side-effects).

memoize is a classic cache and is best used if the calculation is expensive and the set of input arguments is not excessive. There are other caching options in the clojure-verse, that allow better configurations (e.g. they are not unbound).

edited Mar 19, 2020 at 16:15

answered Mar 19, 2020 at 16:11

cfrick

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3 Comments

Lee Over a year ago

delay also caches exceptions unlike memoize.

Esteban Ginocchio Silva Over a year ago

What do you mean by unbound cache?

cfrick Over a year ago

@Estebangs it has no limit; the items in it have no life-time. If you use above function like (run! myinc (range)) your application will eventually run out of memory.

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