Rony Steelandt wrote:
>> Paolo Pantaleo wrote:
>>
>>> I have a function
>>>
>>> def f(the_arg):
>>> ...
>>>
>>> and I want to state that the_arg must be only of a certain type
>>> (actually a list). Is there a way to do that?
>>
>>
>> Yes and no. You can ensure that the passed object is a list, by
>> calling e.g.
>>
>> def f(arg):
>> if not isinstance(arg, list):
>> raise "Not a list!"
>>
>>
>> Alternatively, you can just use it as an iterable - and the exception
>> will
>> come from arg not being iterable.
>>
>> But what you can't do is make python complain about this:
>>
>> def f(arg):
>> for e in arg:
>> print e
>>
>>
>> f(100)
>>
>> before actually calling f. It will always fail at runtime.
>>
>> Diez
>
>
> What about
> def f(arg):
> if type(arg)=='list':
FWIW, type(<some_type>) returns a type object, not a string. So your
test will always fail. A right way to write this is:
if type(arg) is type([]):
...
> #do something
Usually a very bad idea. It defeats the whole point of duck-typing. In
most cases, you don't care about the concrete class - all you want is an
object that implements an (implied) interface.
NB : I say 'usually' because there are a very few cases where testing
the concrete class can make sens - but there again, better to use
isinstance() than type().
--
bruno desthuilliers
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p in '[EMAIL PROTECTED]'.split('@')])"
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