Tail recursion and fold-left

fold-left has this basic recursion:

(fold-left f init ())      = init
(fold-left f init (a . d)) = (fold-left f (f init a) d)

A straightforward implementation of this is

(defun fold-left (f init list)
  (if (null list)
      init
      (fold-left f (funcall f init (car list)) (cdr list))))

The straightforward implementation uses a slightly more space than necessary. The call to f occurs in a subproblem position, so there the stack frame for fold-left is preserved on each call and the result of the call is returned to that stack frame.

But the result of fold-left is the result of the last call to f, so we don't need to retain the stack frame for fold-left on the last call. We can end the iteration on a tail call to f on the final element by unrolling the loop once:

(defun fold-left (f init list)
  (if (null list)
      init
      (fold-left-1 f init (car list) (cdr list))))

(defun fold-left-1 (f init head tail)
  (if (null tail)
      (funcall f init head)
      (fold-left-1 f (funcall f init head) (car tail) (cdr tail))))

There aren't many problems where this would make a difference (a challenge to readers is to come up with a program that runs fine with the unrolled loop but causes a stack overflow with the straightforward implementation), but depending on how extreme your position on tail recursion is, this might be worthwhile.