Compose example in Paul Graham's ANSI Common Lisp

Question

Can anybody explain an example in Paul Graham\'s ANSI Common Lisp page 110?

The example try to explain the use &rest and lambda to create functional programming

Answer 1

destructuring-bind combines destructors with binding. A destructor is a function that lets you access a part of a data structure. car and cdr are simple destructors to extract the head and tail of a list. getf is a general destructor framework. Binding is most commonly performed by let. In this example, fns is (#'list #'round #'sqrt) (the arguments to compose), so (reverse fns) is (#'sqrt #'round #'list). Then

(destructuring-bind (fn1 . rest) '(#'sqrt #'round #'list)
  ...)

is equivalent to

(let ((tmp '(#'sqrt #'round #'list)))
  (let ((fn1 (car tmp))
        (rest (cdr tmp)))
    ...))

except that it doesn't bind tmp, of course. The idea of destructuring-bind is that it's a pattern matching construct: its first argument is a pattern that the data must match, and symbols in the pattern are bound to the corresponding pieces of the data.

So now fn1 is #'sqrt and rest is (#'round #'list). The compose function returns a function: (lambda (&rest args) ...). Now consider what happens when you apply that function to some argument such as 4. The lambda can be applied, yielding

(reduce #'(lambda (v f) (funcall f v))
            '(#'round #'list)
            :initial-value (apply #'sqrt 4)))

The apply function applies fn1 to the argument; since this argument is not a list, this is just (#'sqrt 4) which is 2. In other words, we have

(reduce #'(lambda (v f) (funcall f v))
            '(#'round #'list)
            :initial-value 2)

Now the reduce function does its job, which is to apply #'(lambda (v f) (funcall f v)) successively to the #'round and to #'list, starting with 2. This is equivalent to

(funcall #'list (funcall #'round 2))
→ (#'list (#'round 2))
→ '(2)