"Steven D'Aprano" <st...@remove-this-cybersource.com.au> wrote in message
news:4c3aedd5$0$28647$c3e8...@news.astraweb.com...
On Mon, 12 Jul 2010 09:48:04 +0100, bart.c wrote:
That's interesting. So in Python, you can't tell what local variables a
function has just by looking at it's code:
def foo(day):
if day=="Tuesday":
x=0
print ("Locals:",locals())
#foo("Monday")
Does foo() have 1 or 2 locals?
That's easy for CPython: it prepares two slots for variables, but only
creates one:
foo("Monday")
('Locals:', {'day': 'Monday'})
foo.func_code.co_varnames
('day', 'x')
foo.func_code.co_nlocals
2
So, the question is, is x a local variable or not? It's not in locals,
but the function clearly knows that it could be.
So Alf P.S. could be right; x exists, but Python pretends it doesn't until
it's assigned to.
That might explain some of the
difficulties of getting Python implementations up to speed.
I'm not quite sure why you say that.
Only that an implementation might be obliged to keep these various tables
updated during function execution, instead of just at entry and exit.
--
Bartc
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