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To: python-list@python.org
From: Peter Otten <__peter__@web.de>
Subject: Re: Guarding arithmetic
Date: Thu, 23 Aug 2012 12:11:21 +0200
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Mark Carter wrote:

> Suppose I want to define a function "safe", which returns the argument
> passed if there is no error, and 42 if there is one. So the setup is
> something like:
> 
> def safe(x):
>    # WHAT WOULD DEFINE HERE?
> 
> print safe(666) # prints 666
> print safe(1/0) # prints 42
> 
> I don't see how such a function could be defined. Is it possible?

1/0 is evaluated before safe() is called. Therefore safe() has no chance to 
catch the exception. You have to move the evaluation into the safe() 
function:

>>> def safe(deferred, default=42, exception=Exception):
...     try:
...             return deferred()
...     except exception:
...             return default
... 
>>> print safe(lambda: 666)
666
>>> print safe(lambda: 1/0)
42