scheme – Example of apply other than + in Racket
scheme – Example of apply other than + in Racket
Just like (apply + (list 1 2 3 4)) is equivalent to (+ 1 2 3 4), (apply println (list a b c)) is equivalent to (println a b c). However thats not a valid use of println. As the error message suggests, the second argument should be an output port, not a string.
To correctly call println using apply, you must supply a valid argument list, such as:
(apply println (list a))
;; or
(apply println (list a (current-output-port))
Heres some examples of using apply with functions that take an arbitrary number of arguments and will thus always work regardless of the length of the list:
(apply * (list 1 2 3 4)) ; 24
(apply list (list 1 2 3 4)) ; (list 1 2 3 4) ¹
(apply set (list 1 2 3 4)) ; (set 1 2 3 4)
¹ Well, you wouldnt really use apply with list as that just gives you back the same list (or rather a copy of it), but you can.
If you want to apply a procedure eg. (proc a b c) but the arguments are supplied in a list. You could do (proc (car l) (cadr l) (caddr l)) but you can also just do (apply proc l). If you have additional arguments you want in front you can even do (apply proc 5 l) and it is the same as (proc 5 (car l) (cadr l) (caddr l)). apply takes any number of arguments but the very last needs to be a list of the final arguments.
Its important to understand that apply is not a fold. If you have a procedure that takes exactly one argument then you can apply that with a list of exactly one element or else the procedure will fail..
(cons 1 2) ; ==> (1 . 2)
(apply cons (1 2)) ; ==> (1 . 2)
(cons 1 2 3) ; ==> ERROR cons: arity mismatch
(apply cons (1 2 3)); ==> ERROR cons: arity mismatch
apply is the opposite of a rest argument:
(define (proc arg1 . argn)
(apply list arg1 argn))
(proc 1 2 3 4) ; ==> (1 2 3 4)
It is not a fold
;; a two arity add
(define (add a b)
(+ a b))
(apply add (1 2)) ; ==> 3
(apply add (1 2 3)) ; ==> ERROR add: arity mismatch
;; foldl is a fold
(foldl add 0 (1 2 3)) ; ==> 6