python exception handling

Question

In the below python the message RSU is not supported on single node machine** is not getting printed. can anyone help please??

#! /usr/bin/env python
import sys
class SWMException(Exception):
 def __init__(self, arg):
 print "inside exception"
 Exception.__init__(self, arg)
class RSUNotSupported(SWMException):
 def __init__(self):
 SWMException.__init__(self, "**RSU is not supported on single node machine**")
def isPrepActionNeeded():
 if 1==1:
 raise RSUNotSupported()
try:
 isPrepActionNeeded()
except:
 sys.exit(1)

Answer 1

It is not printed, because you're even not trying to print it :) Here:

try:
 isPrepActionNeeded()
except RSUNotSupported as e:
 print str(e)
 sys.exit(1)

• mandeep: Exception.__init__(self, arg) the arg passed here. what is that for??
• Zaur Nasibov: Well, you do have to keep the "**RSU..." message somewhere, right? Your exception inherits from Exception class, thus Exception.__init__(self, message) initializes the superclass and passes the argument (the message) to it. Now, your exception contains that message, and calling __str__() of an RSUNotSupported class instance (or str( ) on that instance) returns the message.

Answer 2

Because you handle the exception with your try/except clause.

Answer 3

Change the last two lines to:

except Exception as e:
 print e
 sys.exit(1)

I use just Exception here to keep this the equivalent of a bare except:. You really should use RSUNotSupported so you don't hide other types of errors.

• mandeep: Exception.__init__(self, arg) what is the arg argument in this init call for?? it doesnt print the message
• agf: It's to make the message part of the exception, so you can print it yourself later if you want.