​​​​​​​​​​​​​​​​​​​​​Applying a symbol as a procedure

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2

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 +?

scheme

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edited Nov 21, 2009 at 12:57

asked Nov 21, 2009 at 12:32

Yuval Adam

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3 Answers

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3

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)

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answered Dec 2, 2009 at 16:34

Davorak

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0

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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edited Nov 26, 2009 at 21:20

answered Nov 26, 2009 at 12:21

Ben Fitzgerald

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• @Ben - you missed the part where the + is quoted, hence, it is not directly a procedure until it is evaluated. 

  – Yuval Adam

   Nov 26, 2009 at 15:17
• ah yes, see that now. Thanks Yuval. 

  – Ben Fitzgerald

   Nov 26, 2009 at 21:18

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8

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. :)

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answered Nov 21, 2009 at 12:36

Lucas Jones

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• So simple, dunno how I missed that out. Thx! :) 

  – Yuval Adam

   Nov 21, 2009 at 12:39

• 1

  If you want it to work with any environment you should probably use (eval '+ (null-environment 5)) 

  – Joe D

   Dec 5, 2010 at 9:39