Scheme Quickies Answers!
(define (fraggle f) 'rock)
(define four 4)
(define (five) 5)
What is the variable 'fraggle' bound to?
ANSWER: a procedure which takes in any kind of argument and returns the word 'rock'
What is the variable four bound to?
ANSWER: the number 4
What is the variable five bound to?
ANSWER: a procedure which takes no arguments and returns 5
What will Scheme evaluate these expressions to?
(+ '6 1 '7) 14
('+ '6 '1 '7) Error because '+ is not a procedure
(/ 3) 0.33333333
(first 3/6) Error because 3/6 is not a variable
(first '3/6) 3
(bf 3/6) Error because 3/6 is not a variable
(bf 00012) 2
(bf '00012) 2
(bf '10012) "0012"
(se '(fraggle fraggle) #t) Error #t cannot be put into a sentence
(bf (se '(fraggle fraggle) (fraggle fraggle))) (fraggle rock)
(se (fraggle fraggle) (fraggle fraggle)) (rock rock)
(se '((fraggle fraggle) (fraggle fraggle))) Error because sentences cannot have nested sentences
five #[closure...] a procedure
'five five
(five) 5
'(five) (five)
''(five) '(five)
'('five) ('five)
(four) Error because four is not a procedure
(= 1 '1) #t
(= five 5) Error five is not a number
(= four 4) #t
(if (2) 'yes 'no) Error 2 is not a procedure
(if 0 'yes (/ 1 0)) yes
(let ((a 3)) a) 3
(let ((four 5)) (+ four (five))) 10
(member? 'b (or 'red 'blue)) #f
(>= 4 3 2 2 1) #t