Question

Suppose I have a simple symbol:

>  +
+

Is there any way I can apply that symbol as a procedure:

> ((do-something-with  +) 1 2)
3

So that + is evaluated to the procedure +?

Answer 1

I m not 100% sure, but would:

((eval  +) 1 2)

work? I m not sure if you need to specify the environment, or even if that works - I m a Scheme noob. :)

Answer 2

Lucas s answer is great. For untrusted input you can make a white list of allowed symbols/operators.

(define do-something (lambda (op)
                       (cond
                         ((equal? op `+) +)
                         ((equal? op `-) -)
                         ((equal? op `*) *)
                         ((equal? op `/) /)
                         ((equal? op `^) ^))))

((do-something `+) 1 2)

Answer 3

Newbie too so hope I ve understood your question correctly...

Functions are first class objects in scheme so you don t need eval:

1 ]=> (define plus +)

;Value: plus

1 ]=> (plus 2 3)

;Value: 5

HTH

Update: Ignore this and see the comments!

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