Question

In SICP exercise 2.26, this Scheme code is given:

(define x (list 1 2 3))
(define y (list 4 5 6))

Then this cons call is given:

(cons x y)

I expected a pair of lists would result, ((1 2 3) (4 5 6)) but the interpreter gives,
((1 2 3) 4 5 6)
…a list with 4 elements, the first being a list. Why is y treated differently? I’ve tried looking up other SICP answers for an explanation, but couldn’t find something satisfactory. So could any Scheme/Lisp experts please shed some light on this aspect of cons? Thanks in advance for any insight.

Answer 1

'((1 2 3) 4 5 6) is actually a pair of lists. Here’s another way to write it:

'((1 2 3) . (4 5 6))

However, the printer avoids dotted pair notation whenever it can, so you get the first representation instead. The rule is:

'(x . (xs ...))
=>
'(x xs ...)

For any x and xs. Here, your x = '(1 2 3) and xs = '(4 5 6), so you get ((1 2 3) 4 5 6).

---

To see how cons and dotted-pair notation is related, let’s shorten the problem to just '(1) and '(6). The lowest level way to build a pair of them is this:

(cons (cons 1 '()) (cons 6 '()))

Here, '() is nil, or the empty list. If we translate this literally to dotted-pair notation, we get this:

'((1 . ()) . (6 . ()))

But because the printer collapses dotted-pair notation whenever possible, you get this instead:

'((1 . ()) . (6 . ()))
=>
'((1) . (6))    ; <-- x=1, xs=nothing; x=6, xs=nothing
=>
'((1) 6) ; <-- x=1, xs=6