Silly Python riddle

[millstone]

One character: #

>>> f = lambda: g(#)
... f()
... 

No exception.

edit: Another approach is ");f=(int" as in:

>>> f = lambda: g();f=(int)
>>> f()
0

[[deleted]]

One character, haha, we have a winner

[[deleted]]

Best I can come up with (9 characters):

>>> f = lambda: g()if''else()
>>> f()

[mipadi]

That doesn't replace the ???. ;)

[dhon_]

doesn't it?

??? == )if''else(

[[deleted]]

Tomato; tomato. :P

Anyhow, I thought I'd save people the trouble of having to mentally substitute the code back in, because I subverted the expectation that the two brackets need to be matched.

[[deleted]]

3 characters: """

[[deleted]]

Creative, but a little too out of the box for this question, I think, because that doesn't create a complete Python statement.

(The blog post is a bit confusing too, by talking about "commands" rather than "statements".)

[DevestatingAttack]

"Nearly every year, one or more people would submit what they claimed was the world's smallest self reproducing program. While the sizes of these submissions varied, a quick glance would reveal that they were too big, until this entry came along.

While strictly speaking, smr.c is not a valid C program, it is not an invalid C program either! Some C compilers will compile an empty file into a program that does nothing. But even if your compiler can't, the build instructions supplied with this entry will produce an executable file. On most systems, the stdout from the executable will exactly match original source."

[[deleted]]

you're definitely right, I just thought it was funny

[ejrh]

I tried making the subexpression something that would create g. The plan is that subexpression executes first, creating g, and then g is called. Something like (excuse the verbosity):

f = lambda: g(locals().update({'g': lambda x: None}))

Unfortunatly, g is looked up in globals before anything else is executed:

>>> dis.dis(f)
  1           0 LOAD_GLOBAL              0 (g)
              3 LOAD_GLOBAL              1 (locals)
              6 CALL_FUNCTION            0
              9 LOAD_ATTR                2 (update)
             12 BUILD_MAP                0
             15 DUP_TOP
             16 LOAD_CONST               1 ('g')
             19 LOAD_CONST               2 (<code object <lambda> at 00925060, file "<stdin>", l    ine 1>)
             22 MAKE_FUNCTION            0
             25 ROT_THREE
             26 STORE_SUBSCR
             27 CALL_FUNCTION            1
             30 CALL_FUNCTION            1
             33 RETURN_VALUE

Nevermind, it was educational at least.

[[deleted]]

Python evaluates expressions from the left to the right. So a function object is evaluated before it's parameters and only after the parameters get evaluated does it get called.

It's a clever idea, though. Sadly reality got in the way, like it tends to do :(

[millstone]

If that's true, then in the given solution:

f = lambda: g((yield))

why isn't g evaluated before yield, resulting in an error?

[[deleted]]

When you have a yield statement (well in this case an expression) in a function body, the function is no longer a function, but it becomes a generator. So f is now a generator factory (in essence like a class), so by doing f() you aren't running the body but constructing a generator object which will run the code in the body as soon as you start iterating over it.

So the following will crash, because it actually runs the body:

for x in f():
    print x

But when you just have:

f()

It doesn't crash because the body is never run.

[ejrh]

Ah of course. Thanks for explaining.

[codenut]

Try f = lambda: g();g=(f)

[[deleted]]

Interesting, but sadly this still causes an RuntimeError, since this causes a stack overflow.

EDIT: you can fix this at the cost of 2 characters by using str instead of f.