In SICP exercise 2.26 using DrScheme, why does cons return a list, instead of a pair of lists?

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:

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Answer 1

hey, i think you could think of it in this way;

whenever there is a nil, there must be a pair of parenthesis, as follow:

(cons 1 (cons 2 nil))--> (list 1 2)

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

1.(cons x y)--> (cons (cons 1 (cons 2 (cons 3 nil))) (cons 4 (cons 5 (cons 6 nil)))) here, the first nil stands for an end of a pair which could be expressed by parenthesis; whereas the second nil stands for the end of the whole pair which use another pair of parenthesis; so, ((1 2 3) 4 5 6)

2.(list x y)-> (cons x (cons y nil); as we know the x contain a nil, so it must be (1 2 3); the second part contains two nils, so ((1 2 3) (4 5 6));

the inner most nil means the outer most parenthesis;

Hope it can help.