Checking if a variable is a dictionary
guillermo.listas at googlemail.com
Sun Mar 9 14:58:15 CET 2008
Mamma mia! My head just exploded. I've seen the light.
So you only need to ·want· to have a protocol? That's amazing... Far
beyond the claim that Python is easy. You define protocols in writing
basically! Even my grandma could have her own Python protocol.
Okay, so I think I know where's the catch now -- you must rely on the
fact that the protocol is implemented, there's no way to enforce it if
you're expecting a parrot-like object. You'd try to call the speak()
method and deal with the error if there's no such method?
Thanks a lot!
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