Higher order programming with Lisp: Passing a function to mapcar?

Question

I'm just learning ANSI Common Lisp (using clisp on a Win32 machine) and I was wondering if mapcar could use a function passed in as a formal argument? Please see the following:

(defun foo (fn seq) 
    (mapcar #'fn seq))

This would, in my opinion, provide a greater level of flexibility than:

(defun mult (i)
    (* i 2))

(defun foo ()
    (mapcar #'mult '(1 2 3)))

Answer 1

This is absolutely possible! You're almost there. You just bumped into Common Lisp's dual namespaces, which can take a lot of getting used to. I hope I can say a thing or two to make Common Lisp's two namespaces a bit less confusing.

Your code is almost correct. You wrote:

(defun foo (fn seq) 
    (mapcar #'fn seq))

But, what is that trying to do? Well, #' is shorthand. I'll expand it out for you.

(defun foo (fn seq) 
    (mapcar (function fn) seq))

So, #'symbol is shorthand for (function symbol). In Common Lisp -- as you seem to know -- symbols can be bound to a function and to a variable; these are the two namespaces that Lispers talk so much about: the function namespace, and the variable namespace.

Now, what the function special form does is get the function bound to a symbol, or, if you like, the value that the symbol has in the function namespace.

Now, on the REPL, what you wrote would be obviously what you want.

(mapcar #'car sequence)

Would map the car function to a sequence of lists. And the car symbol has no variable binding, only a function binding, which is why you need to use (function ...) (or its shorthand, #') to get at the actual function.

Your foo function doesn't work because the function you pass it as an argument is being bound to a symbol as a variable. Try this:

(let ((fn #'sqrt))
    (mapcar #'fn '(4 9 16 25)))

You might have expected a list of all those numbers' square roots, but it didn't work. That's because you used let to bind the square root function to fn as a variable. Now, try this code:

(let ((fn #'sqrt))
    (mapcar fn '(4 9 16 25)))

Delightful! This binds the square root function to the fn symbol as a variable.

So, let's go revise your foo function:

(defun foo (fn seq) 
    (mapcar fn seq))

That will work, because fn is a variable. Let's test it, just to make sure:

;; This will not work
(foo sqrt '(4 9 16 25))
;; This will work
(foo #'sqrt '(4 9 16 25))

The first one didn't work, because the square root function is bound to sqrt in the function namespace. So, in the second one, we grabbed the function from the symbol, and passed it to foo, which bound it to the symbol fn as a variable.

---

Okay, so what if you want to bind a function to a symbol in the function namespace? Well, for starters, defun does that, permanently. If you want it to be temporary, like a let binding, use flet. Flet is, in my opinion, stupid, because it doesn't work exactly like let does. But, I'll give an example, just so you can see.

(flet ((fn (x) (sqrt x)))
    (mapcar fn '(4 9 16 25)))

will not work, because flet didn't bind the function to the symbol in the variable namespace, but in the function namespace.

(flet ((fn (x) (sqrt x)))
    (mapcar #'fn '(4 9 16 25)))

This will do what you expect, because flet bound that function to the symbol fn in the function namespace. And, just to drive the idea of the function namespace home:

(flet ((fn (x) (sqrt x)))
    (fn 16))

Will return 4.

Answer 2

Sure, you can do this:

(defun foo (fn)
  (mapcar fn '(1 2 3)))

Examples:

(foo #'(lambda (x) (* x 2)))
(foo #'1+)
(foo #'sqrt)
(foo #'(lambda (x) (1+ (* x 3))))