Why are closures broken within exec?

closures, exec, python

Solution

When you pass a string to `exec` or `eval`, it compiles that string to a code object before considering globals or locals. So when you say:

eval('lambda: a', ...)

it means:

eval(compile('lambda: a', '<stdin>', 'eval'), ...)

There's no way for `compile` to know that `a` is a freevar, so it compiles it to a global reference:

>>> c= compile('lambda: a', '<stdin>', 'eval')
>>> c.co_consts[0]
<code object <lambda> at 0x7f36577330a8, file "<stdin>", line 1>
>>> dis.dis(c.co_consts[0])
  1           0 LOAD_GLOBAL              0 (a)
              3 RETURN_VALUE

Therefore to make it work you have to put `a` in the globals and not the locals.

Yeah, it's a bit dodgy. But then that's `exec` and `eval` for you I suppose... they're not supposed to be nice.

Problem

In Python 2.6, ``` >>> exec "print (lambda: a)()" in dict(a=2), {} 2 >>> exec "print (lambda: a)()" in globals(), {'a': 2} Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<string>", line 1, in <module> File "<string>", line 1, in <lambda> NameError: global name 'a' is not defined >>> exec "print (lambda: a).__closure__" in globals(), {'a': 2} None ``` I expected it to print `2` twice, and then print a tuple with a single `cell`. It is the same situation in 3.1. What's going on?

Original source