CODEWITHSUNDEEP
elispemacs24
user2879704
user2879704

Reputation:

Difference between cons and list function call in elisp

In elisp,

(cons 1 2) `returns`
(1 . 2)
(list 1 2) `returns`
(1 2)

What is the difference between both the outputs?

Upvotes: 6

Views: 1574

Answers (3)

Vatine
Vatine

Reputation: 21288

If you have two things, A and B, you have as a given that (eq (car (cons A B)) A) and (eq (cdr (cons A B)) B). That is, cons constructs a "cons cell" (a pair) with two parts, the car part and the cdr part.

By convention, a pair that consists of an atom and the empty list is considered to be a list. So, (cons 2 nil) is (2). By that same token, (list 1 2) returns a similar structure to (cons 1 (cons 2 nil)) (that is, (1 2)).

Upvotes: 1

Maher
Maher

Reputation: 294

the job of cons is to add the first element to the second element so, in your first example (cons 1 2) the 2 is an element so adding 1 to it will create a pair (1 . 2)

the second example creates a list with 1 and 2 as element of that list

if it was like (cons 1 (list 2)) now cons will find that the second parameter is a list and the result will be (1 2)

Upvotes: 0

Barmar
Barmar

Reputation: 782683

(cons 1 2) creates a single cons cell like this:

---------  
| 1 | 2 |   
---------

Lists in Lisp are a chain of cons cells. Each cell's car is a list element, its cdr points to the next cell in the chain; the last one's cdr points to the special symbol nil. So (list 1 2) creates a chain of cons cells like this:

--------|   --------|
| 1 | ----->| 2 | ----> nil
--------|   --------|

It's equivalent to (cons 1 (cons 2 nil))

Upvotes: 4

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