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http://mail.python.org/pipermail/pyt...er/233435.html

why isnt it printing a in the second(second here, last one in OP)
example before complaining?

def run():
a = 1
def run2(b):
a = b
print a
run2(2)
print a
run()

def run():
a = 1
def run2(b):
print a
a = b
run2(2)
print a
run()

globalrev wrote:
> http://mail.python.org/pipermail/pyt...er/233435.html
>
> why isnt it printing a in the second(second here, last one in OP)
> example before complaining?
>
> def run():
> a = 1
> def run2(b):
> a = b
> print a
> run2(2)
> print a
> run()
>
> def run():
> a = 1
> def run2(b):
> print a
> a = b
> run2(2)
> print a
> run()
> --
> http://mail.python.org/mailman/listinfo/python-list
>
If you had told us what error you got, I would have answered you hours
ago. But without that information I ignored you until is was
convenient to run it myself. Now that I see no one has answered, and I
have time to run your examples, I see the error produced is

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 6, in run
File "<stdin>", line 4, in run2
UnboundLocalError: local variable 'a' referenced before assignment

and its obvious what the problem is.

In run2 (of the second example), The assignment to a in the line "a=b"
implies that a *must* be a local variable. Python's scoping rules say
that if "a" is a local variable anywhere in a function, it is a local
variable for *all* uses in that function. Then it's clear that "print
a" is trying to access the local variable before the assignment gives it
a value.

You were expecting that the "print a" pickups it the outer scope "a" and
the assignment later creates a local scope "a", but Python explicitly
refuses to do that.

Gary Herron