SICP Exercise 2.26: Combining Lists

Suppose we define x and y to be two lists:

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

What result is printed by the interpreter in response to evaluating each of the following expressions:

(append x y)

(cons x y)

(list x y)

Let's fire up our interpreter and find out...

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

In the first case we're using append which takes two lists and produces a new list containing all of the elements from both lists in order.

In the second case we're using cons, which creates a pair using its two arguments. As the second argument is a list the resulting structure will also be a list... and as the first argument is a list the first element of the list will be that list.

In the final case we're using list, which creates a list containing all of its arguments. We have two arguments, both of which are lists, so the resulting structure is a two-element list, each element of which is a list.