What woud the interpreter print in response to evaluating each of the following expressions?
(list 'a 'b 'c)
(list (list 'george))
(cdr '((x1 x2) (y1 y2)))
(cadr '((x1 x2) (y1 y2)))
(pair? (car '(a short list)))
(memq 'red '((red shoes) (blue socks)))
(memq 'red '(red shoes blue socks))(list 'a 'b 'c)
; (a b c)
(list (list 'george))
; ((george))
(cdr '((x1 x2) (y1 y2)))
; ((y1 y2))
(cadr '((x1 x2) (y1 y2)))
; (y1 y2)
(pair? (car '(a short list)))
; false
(memq 'red '((red shoes) (blue socks)))
; false
(memq 'red '(red shoes blue socks))
; (red shoes blue socks)Two lists are said to be equal? if they contain equal elements
arranged in the same order. For example,
(equal? '(this is a list) '(this is a list))is true, but
(equal? '(this is a list) '(this (is a) list))is false. To be more precise, we can define equal? recursively in
terms of the basic eq? equality of symbols by saving that a and b
are equal? if they are both symbols and the symbols are eq?, or if
they are both list such that (car a) is equal? to (car b) and
(cdr a) is equal? (cdr b). Using this idea, implement equal?
as a procedure.
(define (equal? x y)
(cond ((null? x) (null? y))
((null? y) (null? x))
((eq? (car x) (car y)) (equal? (cdr x) (cdr y)))
(else false)))Eva Lu Ator types to the interpreter the expression
(car ''abracadabra)To her supprise, the interpreter prints back quote. Explain.
If we translate the expression ''abracadabra to use the quote
special form instead of ', we get
(car (quote (quote abracadabra)))The second inner quote is not evaluated but treated as a symbol, so
that the expression is getting the car of the list
(quote abracadabra) which is quote.