How do I handle an unspecified number of parameters in Scheme?

function scheme variadic-functions variadic
John picture John · Sep 30, 2012 · Viewed 19.8k times · Source

For example ((fn-stringappend string-append) "a" "b" "c") I know how to handle this (f x y z). But what if there's an unknown number of parameters? Is there any way to handle this kind of problem?

Answer

Óscar López picture Óscar López · Sep 30, 2012

In Scheme you can use the dot notation for declaring a procedure that receives a variable number of arguments (also known as varargs or variadic function):

(define (procedure . args)
  ...)

Inside procedure, args will be a list with the zero or more arguments passed; call it like this:

(procedure "a" "b" "c")

As pointed out by @Arafinwe, here's the equivalent notation for an anonymous procedure:

(lambda args ...)

Call it like this:

((lambda args ...) "a" "b" "c")

Remember that if you need to pass the parameters in a list of unknown size to a variadic function you can write it like this:

(apply procedure '("a" "b" "c"))
(apply (lambda args ...) '("a" "b" "c"))

UPDATE:

Regarding the code in the comments, this won't work as you intend:

(define (fp f)
  (lambda (.z)
    (f .z)))

I believe you meant this:

(define (fp f)
  (lambda z
    (apply f z)))

With a bit of syntactic sugar the above procedure can be further simplified to this:

(define ((fp f) . z)
  (apply f z))

But that's just a long way for simply writing:

(apply f z)

Is this what you need?

(apply string-append '("a" "b" "c"))

Because anyway that's equivalent to the following:

(string-append "a" "b" "c")

string-append already receives zero or more arguments (at least, that's the case in Racket)

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