globalrev wrote:
http://mail.python.org/pipermail/python-list/2003-October/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()
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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
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