了解多方法调度

Question

我能理解这是如何工作的:

(defmulti area :Shape)
(defmethod area :B [x] (println "Rect"))
(defmethod area :C [x] (println "Circle"))
(defmethod area :default [x] (do (println "Unknown") (:Shape x)))
(area {:Shape :B})
=> Rect
nil
(area {:Shape :Bb})
=> Unknown
:Bb

但是,通过简单地改变:Shape到:Shap,我停下来理解这是如何分派:

(defmulti area :Shap)
(defmethod area :B [x] (println "Rect"))
(defmethod area :C [x] (println "Circle"))
(defmethod area :default [x] (do (println "Unknown") (:Shap x)))
(area {:Shap :B})
=> Unknown
:B
(area {:Shap :C})
=> Unknown
:C

使用逆时针插件的Eclipse上的Clojure 1.5.1

Answer 1

Multimethods不喜欢使用新的调度函数重新定义.这是一个颇具争议的特征,但事情就是这样.要更改多方法的调度函数,必须首先将其定义为非多方法(例如,为nil).

(defmulti area :Shape)

(def area nil)

(defmulti area :Shap)