How to pass arithmetic operators as function parameters in LISP?

Question

For example, I want to define a function like this:

(defun operation (op)
  (op 3 7))

But the lisp compiler complains about code like this: (operation +)

Is there a way to pass arithmetic operator as function parameters?

possible duplicate of how do I use a function as a variable in lisp? — Joshua Taylor
– Joshua Taylor, Commented Sep 27, 2013 at 15:59
(That's actually not a great duplicate, but a better answer to it would be a good answer for this question.) — Joshua Taylor
– Joshua Taylor, Commented Sep 27, 2013 at 16:04

Neil Forrester · Accepted Answer · 2013-09-27 15:46:28Z

5

There are two categories of Lisp dialects: Lisp-1 and Lisp-2. Lisp-1 means that functions and variables share a single namespace. Scheme is a Lisp-1. Lisp-2 means that functions and variables have different namespaces. Common Lisp is a Lisp-2. If you want to pass a function named a as an argument to another function, you must refer to it as #'a. If you have a function stored in a variable, you can use the apply function to execute it. You code should work if it is rewritten like this:

(defun operation (op)
  (apply op '(3 7)))

(operation #'+)

answered Sep 27, 2013 at 15:46

Neil Forrester

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2 Comments

Rörd Over a year ago

There's also the funcall function that's like apply but with the arguments expanded instead of collected into a list, so (apply op '(3 7)) can also be written as (funcall op 3 7).

Joshua Taylor Over a year ago

For more on when to use which, see When do you use APPLY and when FUNCALL?

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