一种覆盖'type（）'报告的方法

Question

Is there a way to alter a CLASS OBJECT so that type(object) reports a custom string?

class MyClass(object):
    def __init__(self, t):
        if 'detector' in t:
            my_type_string = "I am set as a detector."
        else:
            my_type_string = "I am set as a broadcaster."

>>> o = MyClass('detector')
>>> type(o)
I am set as a detector.

Answer 1

You should not do that. Instead you should implement two separate classes, both of which inherit from MyClass :

class MyClass(object):
    my_type_string = "I am not set to anything."

    def __str__(self):
        return self.my_type_string

class Detector(MyClass):
    my_type_string = "I am set as a detector."

class Broadcaster(MyClass):
    my_type_string = "I am set as a broadcaster."

>>> o = Detector()
>>> type(o)
__main__.Detector
>>> str(o)
'I am set as a detector.'

And if you want to switch your class depending on a string you provide you could implement a factory that returns the desired object:

def factory(t):
    if 'detector' in t:
        return Detector()
    else:
        return Broadcaster()

>>> o = factory('detector')
>>> type(o)
__main__.Detector

Answer 2

You can, but it is dangerous. type is a builtin, and other places could use it. But if you know what you are doing and why you do it, you can redefine type . but Here be dragons. You've been warned. :

class MyClass(object):
    def __init__(self, t):
        if 'detector' in t:
            self.my_type_string = "I am set as a detector."
        else:
            self.my_type_string = "I am set as a broadcaster."

def type(obj):
    return obj.my_type_string if isinstance(obj, MyClass) else __builtins__.type(obj)

But IMHO, you should not use the builtin type name for that use, but create a specific function:

def my_type(obj):
    return obj.my_type_string if hasattr(obj, 'my_type_string') else str(type(obj))

because now it can consistently return a string