Passing a lambda function to a lambda function in Scheme

Question

So I have a little challenge.

I'm trying to program this here:

Which with lambda calculus simplifies to 12.

I have the following Scheme script:

(
    define double (
        lambda x (
            + (car x) (car x)
        )
    )
)

(display 'Debug)
(newline)

(display (double 6))
(newline)


(
    define getTwelve (
        ((
            (lambda x (
                lambda y (
                    (display y)
                    (newline)
                    (x (x y))
                )
            ))
            double
        ) 3)
    )
)



(display getTwelve)

(newline)
(newline)

Which corresponds to this terminal output:

Debug
12
(3)
application: not a procedure;
 expected a procedure that can be applied to arguments
  given: '(#<procedure:double>)
  arguments...:
   '(3)

So of course I thought 'it's because the parameter of double is a list instead of a number' but car y just changes the last line of the terminal output to 3.

I think I'm interpreting the Racket terminal output wrong, but I don't know in which way. There is however the other possibility that passing a lambda function into a lambda function as a parameter is not allowed.

So my question is: Which one is it?

Don't put a newline and indentation after (. It's very unusual Scheme style. — Barmar
– Barmar, Commented Sep 14, 2018 at 19:17
Noted. I was using it to tell the difference between a definition and an execution on the lambda, but you're completely right. :) — Caveman
– Caveman, Commented Sep 14, 2018 at 19:21

Barmar · Accepted Answer · 2018-09-14 19:27:59Z

2

Usually the parameter list after lambda should be a list of variables. You only use a single variable without a list if you want to allow a variable number of arguments to the procedure, which isn't the case here. If you do this, each variable will be bound to successive arguments, and you don't need to use car to get the value of the argument. So it should be:

(define double
    (lambda (x) 
        (+ x x)))

If you do this with getTwelve you should get the result you expect.

(define getTwelve
  ((((lambda (x)
       (lambda (y)
         (display y)
         (newline)
         (x (x y))))
     double) 3)))

You also had an extra set of parentheses around the body of the innermost lambda.

DEMO

edited Sep 14, 2018 at 19:27

answered Sep 14, 2018 at 19:21

Barmar

789k57 gold badges555 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Passing a lambda function to a lambda function in Scheme

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related