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Guy Steele's FOO language
;; Guy Steele's "FOO" language, as found on the mailing list.
;; ----------------------------------------------------------------------
;; [He said..]
;; If you speak Common Lisp, you might find the following
;; bit of code illuminating. (If you speak Scheme but
;; not Common Lisp, then just delete or ignore all occurrences
;; of the strings "funcall " and "#'").
;; This is an interpreter for a tiny Scheme-like language
;; (I'll call it FOO) with just the following constructs:
;; numeric literals
;; variable names
;; LAMBDA expressions
;; IF
;; function calls
;; and three built-in functions:
;; +
;; *
;; (I took the cheesy way out on defining the functions:
;; I just made undefined variables evaluate to themselves
;; and have the @apply function check for those names.
;; That's okay because the data domain of this language
;; is just numbers---there aren't any operations on symbols.)
;; I spelled the name of each functino with a leading "@" purely
;; to avoid conflict with the Common Lisp functions of the
;; same name.
;; You can evaluate an expression by typing at the Common Lisp
;; top level:
;; (@eval '<expression> '() #'(lambda (x) x))
;; Note two things about this piece of code:
;; (1) Every call from one Common Lisp function to another is
;; a tail-call. In other words, in effect I am not using the
;; Common Lisp stack at all to keep information about the state
;; of the FOO program being interpreted.
;; (2) Every LAMBDA expression in the code of the interpreter
;; is a continuation: it says what to do next when the call to
;; any given @-routine is "finished".
;; (3) Every @-routine takes a continuation "cont" and always
;; finishes either by calling cont (as a tail-call) or by
;; calling another @-routine (as a tail-call).
;; If you keep in mind that "#'" means roughly "allocate (in the heap)
;; a closure for the following LAMBDA expression" and that a closure can
;; refer to lexical variables visible to the LAMBDA expression, you can see
;; that one continuation can know about another, which knows about another,
;; and so on; this chain is sometimes called the "control stack", but in
;; this implementation it's all in the heap.
;; Does this help?
;; --Guy
(defun @eval (exp env cont)
(cond ((numberp exp) (funcall cont exp))
((symbolp exp) (@lookup exp env cont))
((eq (first exp) 'LAMBDA)
(funcall cont (list 'CLOSURE (second exp) (third exp) env)))
((eq (first exp) 'IF)
(@eval (second exp) env
#'(lambda (test)
(@eval (cond (test (second exp)) (t (third exp))) env cont))))
(t (@eval (first exp) env
#'(lambda (fn)
(@evlis (rest exp) env
#'(lambda (args) (@apply fn args cont))))))))
(defun @lookup (name env cont)
(cond ((null env) (funcall cont name))
((eq (car (first env)) name) (funcall cont (cdr (first env))))
(t (@lookup name (rest env) cont))))
(defun @evlis (exps env cont)
(cond ((null exps) (funcall cont '()))
(t (@eval (first exps) env
#'(lambda (arg)
(@evlis (rest exps) env
#'(lambda (args) (funcall cont (cons arg args)))))))))
(defun @apply (fn args cont)
(cond ((eq fn '+) (funcall cont (+ (first args) (second args))))
((eq fn '*) (funcall cont (* (first args) (second args))))
((eq fn 'call/cc)
(@apply (first args) (list (list 'CONTINUATION cont)) cont))
((atom fn) (funcall cont 'UNDEFINED-FUNCTION))
((eq (first fn) 'CLOSURE)
(@eval (third fn) (pairlis (second fn) args (fourth fn)) cont))
((eq (first fn) 'CONTINUATION)
(funcall (second fn) (first args)))
(t (funcall cont 'UNDEFINED-FUNCTION))))
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