Chris Fuller wrote:
class Foo(SyntaxError):
... def __init__(self, a,b,c):
... self.args = (a,b,c)
...
raise Foo(1,2,3)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
__main__.Foo: None
Inheriting from SyntaxError doesn't work! When I create a new exception, I
generally subclass from the built-in exception it most resembles, in case
there was some reason to also catch it via an ancestor. But I'm not sure if
that is really all that useful an idea in practice. How do you folk do it?
What do you mean, "doesn't work"? It looks like it works to me. You get a Foo
exception, exactly as expected. The error message isn't what you expect,
because you're making unwarranted assumptions about SyntaxError and how it works.
In general, when you override a method, you take full responsibility to
perform everything that the superclass method was supposed to do. In this
case, you fail to assign to msg as well as args. It is safer to overload a
message rather than override it:
>>> class Spam(SyntaxError):
... def __init__(self, *args):
... if args:
... args = ("I pity the fool who made a mistake",) +
args[1:]
... super(Spam, self).__init__(*args)
...
>>>
>>> raise Spam('you made a mistake', 1, 2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
__main__.Spam: I pity the fool who made a mistake
Unfortunately, there's no real consistency in what arguments exceptions are
expected to take. The best thing is to read the docs, if they have any, or use
introspection and trial and error to work out what they do.
>>> try:
... raise SyntaxError("you made a mistake")
... except SyntaxError, err:
... pass
...
>>> err.msg
'you made a mistake'
See dir(err) for more; you can use help(SyntaxError) but unfortunately it
isn't very useful.
You probably shouldn't inherit from SyntaxError, since it represents syntax
errors in the Python code being interpreted or compiled. Any syntax error in
your own data structures should be independent of SyntaxError.
--
Steven
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