CODEWITHSUNDEEP
functionvariablesscheme
Leslie G
Leslie G

Reputation: 309

How variables are sent to the functions in scheme?

I'm new to scheme. When I run the following code

(define lst '(1))

(let ((func1 (lambda lst 
               (begin (display lst)
                      lst))))
  (begin (display lst) 
         (func1 lst)))

I got (1)((1))'((1)), which means lst displays as (1) when called in the fourth line, but when send it to the function func1, it becomes ((1)). What exactly happened here?

Upvotes: 1

Views: 64

Answers (2)

Óscar López
Óscar López

Reputation: 236004

In a lambda form, when you write this:

(lambda x <body>)

... you're declaring that x is a list of parameters with zero or more elements. On the other hand, this:

(lambda (x) <body>)

... is stating that x is a single parameter. In the question, this code (the begin is unnecessary for the body part in a lambda):

((lambda lst (display lst) lst) '(1))

... will display and return the list of parameters; if we pass '(1) it will evaluate to '((1)): a list with a single element which happens to be a list.

Surely you intended to do this instead:

((lambda (lst) (display lst) lst) '(1))

... which will display and return the single parameter it receives - in case of passing '(1) the above expression will evaluate to '(1), the parameter itself.

Upvotes: 2

Fred Foo
Fred Foo

Reputation: 363607

(lambda Args E) means: bind the variable-length argument list of this function to Args. E.g.

(define f (lambda args `(got ,(length args) arguments)))
(display (f 'foo 'bar 'baz))

will print (got 3 arguments). If you change the lambda expression to

(lambda (lst) (begin (display lst) lst))
;;;     ^---^

then the function will print, and return, its single argument.

Upvotes: 3

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