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

Question

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 1

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)