Differences between a seq and a list

Question

What's the differences between seqs and lists in Clojure language?

(list [1 2 3]) => ([1 2 3])
(seq [1 2 3]) => ([1 2 3])

These two forms seem to be evaluated as the same results.

Answer 1

First of all, they may seem to be the same, but they're not:

(class (list [1 2 3])) => clojure.lang.PersistentList
(class (seq [1 2 3])) => clojure.lang.PersistentVector$ChunkedSeq

list is usually an implementation, whereas seq is always an abstraction.

The differences between seqs and lists lies in the following three aspects, as pointed out in Clojure Programming

1. getting the length of a seq can be costy:

e.g. from Clojure Programming

(let [s (range 1e6)]
  (time (count s))) => 1000000
; "Elapsed time: 147.661 msecs"

(let [s (apply list (range 1e6))]
  (time (count s))) => 1000000
; "Elapsed time: 0.03 msecs

Because a list always holds a record of its own length, so the operation of counting a list costs constant time. A seq, however, needs to traverse itself to retrieve its count.

2. seqs can be lazy, whereas lists cannot.

 (class (range)) => clojure.lang.LazySeq
 (class (apply list (range))) ;cannot be evaluated
 ; "java.lang.OutOfMemoryError: GC overhead limit exceeded"

3. seqs can be infinite, thus uncountable, whereas lists are always countable.

Also, lists are their own seqs (implementation details):

(class (seq '(1 2 3))) => clojure.lang.PersistentList

One can always create a seq using cons. Check out more information in this post for differences between cons and conj.