Question

在SBCL中，当我定义新的元类

时

CL-USER> (defclass counting-class (standard-class)
   ((counter :initform 0)))
#<STANDARD-CLASS COUNTING-CLASS>

并向GF“make-instance”添加一个方法：

CL-USER> (defmethod make-instance :after ((class counting-class) &key)
   (incf (slot-value class 'counter)))
#<STANDARD-METHOD MAKE-INSTANCE :AFTER (COUNTING-CLASS) {25302219}>

如果我尝试创建实例，我收到错误：

CL-USER> (defclass counted-point () (x y) (:metaclass counting-class))

The class #<STANDARD-CLASS STANDARD-OBJECT> was specified as a
super-class of the class #<COUNTING-CLASS COUNTED-POINT>, but
the meta-classes #<STANDARD-CLASS STANDARD-CLASS> and
#<STANDARD-CLASS COUNTING-CLASS> are incompatible.  Define a
method for SB-MOP:VALIDATE-SUPERCLASS to avoid this error.

现在，如果我添加所需的定义：

CL-USER>  (defmethod sb-mop:validate-superclass ((class counting-class)
                                                 (super standard-class))
            t)
#<STANDARD-METHOD SB-MOP:VALIDATE-SUPERCLASS (COUNTING-CLASS STANDARD-CLASS) {26443EC9}>

有效：

CL-USER> (defclass counted-point () (x y) (:metaclass counting-class))
#<COUNTING-CLASS COUNTED-POINT>

我的问题是：为什么需要这个？

从我的POV中，将计数类声明为标准类的衍生物就足够了，就像我在第一步中所做的那样。

Answer 1

CLOS MOP的validate-superclass规范说明默认方法仅在微不足道的情况下返回t并添加：

在validate-superclass上定义方法需要详细的知识两个类中的每个类的内部协议元对象类。 validate-superclass上返回true的方法对于两个不同类的metaobject类声明它们是兼容。

你可以认为你的validate-superclass是一个你明白自己在做什么的宣言。

顺便说一句，我认为您可以更轻松地定义a class which would count its instances。

PS。 Some implementations also return t in some other cases

sbcl / CLOS为什么我必须在这里添加“validate-superclass”-Method？

1 个答案: