On Sat, Dec 17, 2011 at 6:14 PM, Shai Berger wrote:
> > (Note that these are the only commands that I ran. You're not allowed
> > run any other commands before them.)
> > The riddle: What's the shortest thing you can put instead of *???* so
> > second command would not raise an exception?
> ???= (yield)
> (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
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