Checking if a variable is a dictionary

[Steven D'Aprano]

On Sun, 09 Mar 2008 06:58:15 -0700, Guillermo wrote:

> 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?

That's right. That's called "duck typing" -- if all you want is something 
that quacks like a duck, then it doesn't matter if it actually is a duck 
or not.

Or if you prefer: if it quacks like a duck and swims like a duck, then 
it's close enough to a duck as to make no difference.


Sometimes though, you need to check for a parrot up front. So I'd so this:

try:
    something.speak
except AttributeError:
    # No speak() method, so it can't be a parrot.
    do_something_else()
else:
    # It seems to follow the parrot protocol.
    yummy_goodness = something.speak(5)
    assert "spam" in yummy_goodness.lower()


-- 
Steven