functions, optional parameters

[Chris Angelico]

On Sat, May 9, 2015 at 11:41 AM, Gregory Ewing
<greg.ewing at canterbury.ac.nz> wrote:
> Chris Angelico wrote:
>>
>> How do you know that the function's code
>> object was created when compile() happened, rather than being created
>> when the function was defined?
>
>
> Python 3.4.2 (default, Feb  4 2015, 20:08:25)
> [GCC 4.2.1 (Apple Inc. build 5664)] on darwin
> Type "help", "copyright", "credits" or "license" for more information.
>>>> source = "def f(x = 42): pass"
>>>> code = compile(source, "", "exec")
>>>> c1 = code.co_consts[1]
>>>> c1
> <code object f at 0x53d430, file "", line 1>
>>>> e = {}
>>>> exec(code, e)
>>>> c2 = e['f'].__code__
>>>> c2
> <code object f at 0x53d430, file "", line 1>
>>>> c1 is c2
> True
>
> Is that proof enough for you?

That's what I reached for as my first try, but it's no different from this:

>>> def f(x=42): return x + 1
...
>>> n1 = f()
>>> n2 = f(n1-1)
>>> n1 is n2
True

Clearly in this case, the "x + 1" is getting evaluated at run-time,
and yet the interpreter is welcome to intern the constants. So no, it
isn't proof - it's equally well explained by the code object being
constant.

ChrisA