(let [foo {:x 1}]
(= foo foo))
is fast, because they are identical, and = is fast for things that are
identical.
In general, two maps that are = are often not identical.
If two maps have different numbers of elements, = should quickly return
false because of the way equiv() is implemented for maps, here:
https://github.com/clojure/clojure/blob/master/src/jvm/clojure/lang/APersistentMap.java#L73-L94
There is a check whether the maps are the same size, and if not, return
false without checking all key/value pairs.
If they have equal number of key/value pairs, then it will iterate through
all key/value pairs of one, checking whether the key of one is present in
the other, and if so, whether their associated values are equiv() or not.
Those checks can be fast if the keys are quickly comparable, and the values
are quickly comparable.
If you take one persistent map m1 and remove a few keys and/or add a few
keys to produce m2, then all of the unchanged keys and values they have in
common will be identical, and very quickly comparable. The slowest part of
comparing them would be comparing their non-identical parts.
Andy
On Sat, Jun 27, 2015 at 5:01 AM, Jacob Goodson <submissionfight...@gmx.com>
wrote:
> I was wondering something...
>
> Would (= {:x 1} {:x 1}) be just as fast as (= 1 1) since maps are
> immutable? The idea is that since clojure data structures are just values
> can they be compared much faster than there mutable counter parts? Thanks.
>
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