bug#11328: 24.1.50; Comment in `dired-copy-file-recursive' code

[Drew Adams]

> we can't say also they are really "free" (even if they
> are sort of free) because they are (will be) all
> let-bounded in some places.

Look up the definition of free variable - see the lambda calculus.

It is always about free _occurrences_ of a variable.  Freedom of a variable (not
a great word for it, admittedly) is relative to a given context.

Whether a variable is bound in some outer lambda is not the question.  What
matters is whether it is free in some other (e.g. inner) lambda.

X is a free variable in the inner lambda form here, though it is bound in the
outer one, i.e., in the overall expression.

(funcall
 (funcall
   (lambda (x) (lambda (y) (+ x y)))
   42) ; Fn that increments by 42.
 3) ; Applied to 3 gives 45.

That is equivalent to this:

(let ((x 42))
  (funcall (lambda (y) (+ x y)) 3)

(It's perhaps clearer if you remove all the `funcall' atoms.  That will give you
Lisp 1 syntax ~= lambda calculus syntax.)