Clojure: cons (seq) vs. conj (list)

CollectionsClojureSequence

Collections Problem Overview


I know that cons returns a seq and conj returns a collection. I also know that conj "adds" the item to the optimal end of the collection, and cons always "adds" the item to the front. This example illustrates both of these points:

user=> (conj [1 2 3] 4) ; returns a collection
[1 2 3 4]
user=> (cons 4 [1 2 3]) ; returns a seq
(4 1 2 3)

For vectors, maps, and sets these differences make sense to me. However, for lists they seem identical.

user=> (conj (list 3 2 1) 4) ; returns a list
(4 3 2 1)
user=> (cons 4 (list 3 2 1)) ; returns a seq
(4 3 2 1)

Are there any examples using lists where conj vs. cons exhibit different behaviors, or are they truly interchangeable? Phrased differently, is there an example where a list and a seq cannot be used equivalently?

Collections Solutions


Solution 1 - Collections

One difference is that conj accepts any number of arguments to insert into a collection, while cons takes just one:

(conj '(1 2 3) 4 5 6)
; => (6 5 4 1 2 3)

(cons 4 5 6 '(1 2 3))
; => IllegalArgumentException due to wrong arity

Another difference is in the class of the return value:

(class (conj '(1 2 3) 4))
; => clojure.lang.PersistentList

(class (cons 4 '(1 2 3))
; => clojure.lang.Cons

Note that these are not really interchangeable; in particular, clojure.lang.Cons does not implement clojure.lang.Counted, so a count on it is no longer a constant time operation (in this case it would probably reduce to 1 + 3 -- the 1 comes from linear traversal over the first element, the 3 comes from (next (cons 4 '(1 2 3)) being a PersistentList and thus Counted).

The intention behind the names is, I believe, that cons means to cons(truct a seq)1, whereas conj means to conj(oin an item onto a collection). The seq being constructed by cons starts with the element passed as its first argument and has as its next / rest part the thing resulting from the application of seq to the second argument; as displayed above, the whole thing is of class clojure.lang.Cons. In contrast, conj always returns a collection of roughly the same type as the collection passed to it. (Roughly, because a PersistentArrayMap will be turned into a PersistentHashMap as soon as it grows beyond 9 entries.)


1 Traditionally, in the Lisp world, cons cons(tructs a pair), so Clojure departs from the Lisp tradition in having its cons function construct a seq which doesn't have a traditional cdr. The generalised usage of cons to mean "construct a record of some type or other to hold a number of values together" is currently ubiquitous in the study of programming languages and their implementation; that's what's meant when "avoiding consing" is mentioned.

Solution 2 - Collections

My understanding is that what you say is true: conj on a list is equivalent to cons on a list.

You can think of conj as being an "insert somewhere" operation, and cons as being an "insert at the head" operation. On a list, it is most logical to insert at the head, so conj and cons are equivalent in this case.

Solution 3 - Collections

Another difference is that because conj takes a sequence as the first argument, it plays nicely with alter when updating a ref to some sequence:

(dosync (alter a-sequence-ref conj an-item))

This basically does (conj a-sequence-ref an-item) in a thread-safe manner. This wouldn't work with cons. See the chapter on Concurrency in Programming Clojure by Stu Halloway for more info.

Solution 4 - Collections

Another difference is the behavior of list?

(list? (conj () 1)) ;=> true
(list? (cons 1 ())) ; => false

Solution 5 - Collections

There are dedicated functions in the Tupelo Library to add append or prepend values to any sequential collection:

(append [1 2] 3  )   ;=> [1 2 3  ]
(append [1 2] 3 4)   ;=> [1 2 3 4]

(prepend   3 [2 1])  ;=> [  3 2 1]
(prepend 4 3 [2 1])  ;=> [4 3 2 1]

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiondbyrneView Question on Stackoverflow
Solution 1 - CollectionsMichał MarczykView Answer on Stackoverflow
Solution 2 - CollectionsDaniel YankowskyView Answer on Stackoverflow
Solution 3 - Collectionsuser323818View Answer on Stackoverflow
Solution 4 - CollectionsFredAKAView Answer on Stackoverflow
Solution 5 - CollectionsAlan ThompsonView Answer on Stackoverflow