(+)
(+ 1 2)
(+ 1 2 3)
(*)
(* 2 3 4)
(-)
(modulo 14 2)
(display `(1 2 3 4))
(display 10/20)
(display "\n")
(display 1)
(display "\n")
(display 100.111111)
#(1 2 3 4)
(display #\a)
`(1 2 3 4)
(quote (1 2 3 4))
(1 2 3 4)
The evaluation rules of scheme are as follows
(define a `(+ 10 20 30 40))
a
(eval a)
(car `(1 2 3 4))
(cdr `(1 2 3 4))
(cons 1 `(2 3 4))
Use cons to create:
(cons 1 (cons 2 (cons 3 ())))
; On GL it would be (cons 1 (cons 2 (cons 3 null)) due to dialectal differences
(cons (cons 1 (cons 2 ())) (cons 3 ()))
(cons 1
(cons (cons 2 (cons 3 ())) ())
)
(cons (cons 1 (cons 2 ())) (cons (cons 3 ()) ()))
(cons (cons 1 ()) (cons (cons 2 ()) (cons (cons 3 ()) () )))
(cons (cons (cons 1 ()) () ) (cons (cons 2 ()) (cons 3 ())))
Get 3 out of the following expressions:
(car (car (car `(((3)) (1) 2))))
(caaar `(((3)) (1) 2))
(car (car (cdr `((1) (3) ((2))))))
(caadr `((1) (3) ((2))))
(car (cdr (car `((1 3 2) 5))))
(cadar `((1 3 2) 5))
(eq? "the" "the")
(eq? `t 't)
(equal? "the" "the")
(equal? (cons 1 (cons 2 (cons 3()))) `(1 2 3))
(list 1 2 3 4)
(append `(1 2) `(3 4))
(append `(1 2) `3 `4)
(append `(1 2) `(3) `("s"))
(list 1 2 3)
(list (list 1 2) 3)
(list 1 (list 2 3))
(list (list 1 2) (list 3))
(list (list 1) (list 2) (list 3))
(list (list (list 1)) (list 2 ) 3)