[Python-il] Silly Python riddle

Alon Levy alonlevy1 at gmail.com
Sat Dec 17 18:34:11 IST 2011


cute and interesting.

On Sat, Dec 17, 2011 at 6:27 PM, Ram Rachum <ram at rachum.com> wrote:
> On Sat, Dec 17, 2011 at 6:14 PM, Shai Berger <shai at platonix.com> wrote:
>>
>> >
>> > (Note that these are the only commands that I ran. You're not allowed to
>> > run any other commands before them.)
>> >
>> > The riddle: What's the shortest thing you can put instead of *???* so my
>>
>> > second command would not raise an exception?
>> >
>> >
>> ???= (yield)
>>
>> right?
>>
>> (mailed privately, to avoid ruining the fun...)
>
>
> Yep!!! I just almost finished writing the email to tell everyone that when I
> got your answer.
>
> Congrats for solving the riddle Shai.
>
> So as Shai said, the solution is:
>
>>
>> >>> f = lambda: g((yield))
>> >>> f()
>
>
> Funny, isn't it? I was surprised to see that the `yield` keyword can be used
> in a lambda function.
>
> So when you type `f()`, it just returns a generator. If you'll try to
> exhaust it, an exception will be raised because `g` doesn't exist, but
> that's a new line :)
>
> It's funny that in this case, Python seems to throw away the value of the
> lambda function! As we know, the `yield` keyword actually forms a statement
> whose value is `None`, unless you used the generator's `.send` instead of
> `.next`. So you could also use `.send` to send in whatever value you want
> into the lambda function, and Python will just throw it away. Unless I'm
> missing something.
>
> So that's the only case I can think of where Python completely throws away
> the value of a lambda function.
>
>
> Another funny thing that I learned from this riddle is that when you do a
> function invocation in Python, Python accesses the function before it looks
> at the arguments.
>
> So if were to do:
>
>> adfgadgof(1 / 0)
>
>
> Python will complain about the non-existent function before it even sees the
> division-by-zero.
>
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-- 
Alon Levy


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