在Python中使用__slots__方法的详细教程
正常情况下，当我们定义了一个class，创建了一个class的实例后，我们可以给该实例绑定任何属性和方法，这就是动态语言的灵活性。先定义class：
>>> class Student(object): ... pass ...
然后，尝试给实例绑定一个属性：
>>> s = Student() >>> s.name = 'Michael' # 动态给实例绑定一个属性 >>> print s.name Michael
还可以尝试给实例绑定一个方法：
>>> def set_age(self, age): # 定义一个函数作为实例方法 ... self.age = age ... >>> from types import MethodType >>> s.set_age = MethodType(set_age, s, Student) # 给实例绑定一个方法 >>> s.set_age(25) # 调用实例方法 >>> s.age # 测试结果 25
但是，给一个实例绑定的方法，对另一个实例是不起作用的：
>>> s2 = Student() # 创建新的实例 >>> s2.set_age(25) # 尝试调用方法 Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: 'Student' object has no attribute 'set_age'
为了给所有实例都绑定方法，可以给class绑定方法：
>>> def set_score(self, score): ... self.score = score ... >>> Student.set_score = MethodType(set_score, None, Student)
给class绑定方法后，所有实例均可调用：
>>> s.set_score(100) >>> s.score 100 >>> s2.set_score(99) >>> s2.score 99
通常情况下，上面的set_score方法可以直接定义在class中，但动态绑定允许我们在程序运行的过程中动态给class加上功能，这在静态语言中很难实现。
使用__slots__
但是，如果我们想要限制class的属性怎么办？比如，只允许对Student实例添加name和age属性。
为了达到限制的目的，Python允许在定义class的时候，定义一个特殊的__slots__变量，来限制该class能添加的属性：
>>> class Student(object):
... __slots__ = ('name', 'age') # 用tuple定义允许绑定的属性名称
...
然后，我们试试：
>>> s = Student() # 创建新的实例 >>> s.name = 'Michael' # 绑定属性'name' >>> s.age = 25 # 绑定属性'age' >>> s.score = 99 # 绑定属性'score' Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: 'Student' object has no attribute 'score'
由于'score'没有被放到__slots__中，所以不能绑定score属性，试图绑定score将得到AttributeError的错误。
使用__slots__要注意，__slots__定义的属性仅对当前类起作用，对继承的子类是不起作用的：
>>> class GraduateStudent(Student): ... pass ... >>> g = GraduateStudent() >>> g.score = 9999 Try
除非在子类中也定义__slots__，这样，子类允许定义的属性就是自身的__slots__加上父类的__slots__。
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