Reputation: 53481
I am learning Scheme. What is wrong with the code below?I want to write a program that takes the first function from the list and then applies that to a number?
(define num 3)
;;I want to do something like this which returns 3
((λ (x) x)num)
;;but my functions are in a list so this should return3
((first '((λ (x) x) (λ (x) (* x x)))) num)
Im getting this error for the above code:
procedure application: expected procedure, given: (λ (x) x); arguments were: 3
What does it mean when I get these kinds of output?
When I dont apply anything, I get a nice output.
(first '((λ(x) x)(λ(x) (*x x))))
returns (λ (x) x)
Upvotes: 6
Views: 1124
Reputation: 480
(lambda (x) x) isn't a procedure. It's a form that evaluates to a procedure. People are a bit loose with terminology and often call the lambda form a procedure as a kind of shorthand. “Ceci n'est pas une pipe.”
Upvotes: 2
Reputation: 1417
What is the difference between these expressions?
> (procedure? (lambda (n) n))
#t
> (procedure? (quote (lambda (n) n)))
#f
> (procedure? '(lambda (n) n))
#f
Jay answered it for you but I can't upvote him yet.
Upvotes: 2
Reputation: 8783
You're quoting, with ' the lambda, so it isn't being evaluated.
If you just feed in (λ (x) x) at the prompt, DrScheme shows you #<procedure>
, which means it has actually evaluated the lambda, and given you back a closure. By quoting it, you're giving Scheme just a list of symbols.
If you want to put your functions in a list, you can do:
((first (list (lambda (x) x) (lambda (x) (* x x)))) num)
The quote allows you to produce a list, yes, but one whose contents aren't evaluated. The list function produces a list from all of its arguments, after they've been evaluated.
You could also quasiquote the list, if you like:
((first `(,(lambda (x) x) ,(lambda (x) (* x x)))) num)
Upvotes: 11