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。
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