История изменений
Исправление den73, (текущая версия) :
Вот тебе задача, если нечем заняться:
;;; Problems:
;;;
;;; -- Constraint propagation badly interacts with bottom-up type
;;; inference. Consider
;;;
;;; (defun foo (n &aux (i 42))
;;; (declare (optimize speed))
;;; (declare (fixnum n)
;;; #+nil (type (integer 0) i))
;;; (tagbody
;;; (setq i 0)
;;; :loop
;;; (when (>= i n) (go :exit))
;;; (setq i (1+ i))
;;; (go :loop)
;;; :exit))
;;;
;;; In this case CP cannot even infer that I is of class INTEGER.
;;;
;;; -- In the above example if we place the check after SETQ, CP will
;;; fail to infer (< I FIXNUM): it does not understand that this
;;; constraint follows from (TYPEP I (INTEGER 0 0)).
Исправление den73, :
Вот тебе задача, если нечем заняться:
;;; Problems:
;;;
;;; -- Constraint propagation badly interacts with bottom-up type
;;; inference. Consider
;;;
;;; (defun foo (n &aux (i 42))
;;; (declare (optimize speed))
;;; (declare (fixnum n)
;;; #+nil (type (integer 0) i))
;;; (tagbody
;;; (setq i 0)
;;; :loop
;;; (when (>= i n) (go :exit))
;;; (setq i (1+ i))
;;; (go :loop)
;;; :exit))
;;;
;;; In this case CP cannot even infer that I is of class INTEGER.
;;;
;;; -- In the above example if we place the check after SETQ, CP will
;;; fail to infer (< I FIXNUM): it does not understand that this
;;; constraint follows from (TYPEP I (INTEGER 0 0)).
[/quote]
А обсуждать твой вопрос уже несколько поздно. Исходная версия den73, :
Вот тебе задача, если нечем заняться:
;;; Problems: ;;; ;;; — Constraint propagation badly interacts with bottom-up type ;;; inference. Consider ;;; ;;; (defun foo (n &aux (i 42)) ;;; (declare (optimize speed)) ;;; (declare (fixnum n) ;;; #+nil (type (integer 0) i)) ;;; (tagbody ;;; (setq i 0) ;;; :loop ;;; (when (>= i n) (go :exit)) ;;; (setq i (1+ i)) ;;; (go :loop) ;;; :exit)) ;;; ;;; In this case CP cannot even infer that I is of class INTEGER. ;;; ;;; — In the above example if we place the check after SETQ, CP will ;;; fail to infer (< I FIXNUM): it does not understand that this ;;; constraint follows from (TYPEP I (INTEGER 0 0)).
А обсуждать твой вопрос уже несколько поздно.