在对类MyException（Exception）进行初始化时，初始化self.args的原因是什么？

Question

class DeviceError(Exception):
    def __init__(self,errno,msg):
        self.args = (errno, msg)
        self.errno = errno
        self.errmsg = msg

# Raises an exception (multiple arguments)
raise DeviceError(1, 'Not Responding')

Beazley: pg 88

"it is important to assign a tuple containing the arguments to _ init _() to the attribute self.args as shown.This attribute is used when printing exception traceback messages. If you leave it undefined, users won't be able to see any useful information about the exception when an error occurs."

If i do:

try:
 ....
except DeviceError:
 ....

here self.args is not used since a Traceback is not generated - Correct? if i ignore DeviceError for some reason, then the sys.excepthook() function that is called will need to print a Traceback and will look in self.args - correct? What does it look for? I mean I'm just stuffing random values in a tuple.. how does the default error handler(excepthook function) know how to display errno and msg?

Could someone explain what exactly goes on with self.args and is it used in Python 3.x?

Answer 1

args is used in the __str__ and __repr__ of the base Exception type. Translating from the C source, these are roughly as follows:

def __str__(self):
    return ("" if len(self.args) == 0 else
            str(self.args[0]) if len(self.args) == 1 else
            str(self.args))

def __repr__(self):
    return "%s%r" % (self.__class__.__name__.split('.')[-1], self.args)

You don't need to set args , but it means you don't need to write your own __str__ or __repr__ .

Also, rather than setting args yourself you should pass it to the parent constructor:

class DeviceError(Exception):
    def __init__(self, errno, msg):
        super(DeviceError, self).__init__(errno, msg)
        self.errno = errno
        self.errmsg = msg