Kent Johnson wrote:
> Antoon Pardon wrote:
>> This make me wonder. Would it be possible to do something with
>> metaclasses so that after
>>
>> class SomeClass(MetaClass):
>> ...
>>
>> SomeClass() will be equivalent to MetaClass.__call__(SomeClass)
>
> I think that's already what happens. IIUC type.__call__ implements the
> standard class creation calling __new__ and __init__.
You are a layer of abstraction off.
class SomeClass(MetaClass):
...
creates a subclass of MetaClass, (the superclass of SomeClass).
SomeClass is still an instance of type (assuming new-style classes).
class FunnyType(type):
...
defines a new metaclass, which can affect how instances of FunnyType
behave.
For a ridiculous example:
class FunnyType(type):
def __call__(self, *args, **kwargs):
if 'please' in kwargs and kwargs['please']:
kwargs.pop('please')
return super(FunnyType, self).__call__(*args, **kwargs)
print 'Nope, use your words:', args, kwargs
class SomeClass:
__metaclass__ = FunnyType
def __repr__(self):
print 'Lucky_%s' % id(self)
OK, now:
v = SomeClass(please=True)
print v
works, but
v = SomeClass()
doesn't, nor does:
v = SomeClass(please=False)
Think aboutwhat this code should do before running it, the mental
exercise is fun.
--Scott David Daniels
[EMAIL PROTECTED]
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