Python中type构造函数参数说明

测试代码如下：

 1 class ModelMetaClass(type):
 2     def __new__(cls,name,base,attrs):
 3         logging.info("cls is:"+str(cls))
 4         logging.info("name is:"+str(name))
 5         logging.info("base is:"+str(base))
 6         logging.info("attrs is:"+str(attrs))
 7         return type.__new__(cls,name,base,attrs)
 8         pass
 9     pass
10
11 class Model(dict):
12     __metaclass__ = ModelMetaClass
13     def __init__(self):
14         pass
15     pass
16
17 def main():
18     m=Model()
19
20 if __name__ == ‘__main__‘:
21     main()

测试结果：

1 INFO:root:cls is:<class ‘__main__.ModelMetaClass‘>
2 INFO:root:name is:Model
3 INFO:root:base is:(<type ‘dict‘>,)
4 INFO:root:attrs is:{‘__module__‘: ‘__main__‘, ‘__metaclass__‘: <class ‘__main__.ModelMetaClass‘>, ‘__init__‘: <function __init__ at 0x025C76F0>}

结论就显而易见了。cls是当前类的名字，即类本身。name是通过__metaclass__属性指向ModelMetaClass的类，即要实例化的类，ModelMetaClass拦截了该类的实例化。base是要实例化的类的基类。attrs是要实例化的类的属性集合。