Question

我从面向对象程序员的函数编程一书中得到了这个clojure代码：

(def point {:x 1, :y 2, :__class_symbol__ 'Point})

(def Point
     (fn [x y]
       {:x x,
        :y y
        :__class_symbol__ 'Point}))

(def x :x)
(def y :y)
(def class-of :__class_symbol__)

(def shift
     (fn [this xinc yinc]
       (Point (+ (x this) xinc)
              (+ (y this) yinc))))


(defn add [left right]
  (shift left (x right) (y right)))


(def Triangle
     (fn [point1 point2 point3]
       {:point1 point1, :point2 point2, :point3 point3
        :__class_symbol__ 'Triangle}))


(def right-triangle (Triangle (Point 0 0)
                              (Point 0 1)
                              (Point 1 0)))

(def equal-right-triangle (Triangle (Point 0 0)
                                    (Point 0 1)
                                    (Point 1 0)))

(def different-triangle (Triangle (Point 0 0)
                                  (Point 0 10)
                                  (Point 10 0)))

(defn equal-triangles? [& args]
  (apply = args))

(defn make [klass & args]
  (apply klass args))

我创建了这个函数来检查Triangle伪类的相等性：

(defn equal-triangles? [& args]
  (apply = args))

很明显为什么这个表达式是真的

(= right-triangle right-triangle)

为什么这个表达式不正确也很明显

(equal-triangles? right-triangle different-triangle)

不明显的是为什么这个表达式为真：

(= right-triangle equal-right-triangle)

他们都有Point值，但我认为他们会有所不同，因为我可能仍在考虑实例。

任何人都可以解释为什么最后一个表达式是真的吗？

Answer 1

user=> (= right-triangle equal-right-triangle)
true
user=> (identical? right-triangle equal-right-triangle)
false

user=> (doc =)
-------------------------
... Clojure's immutable data structures define equals() (and thus =) 
as a value, not an identity, comparison.

user=> (doc identical?)
-------------------------
...Tests if 2 arguments are the same object

比较伪类实例时的clojure等式

1 个答案: