why does this work:
def function1():
a = 10
def function2():
print a
function2()
but this does not:
def function1():
a = 10
def function2():
print a
a -= 1
if a>0:
function2()
function2()
I get this error:
UnboundLocalError: local variable 'a' referenced before assignment
The error doesn't seem to be very descriptive of the root problem. Mike explains the messages but that does not explain the root cause.
The actual problem is that in python you cannot assign to closed over variables. So in function2 'a' is read only. When you assign to it you create a new variable which, as Mike points out, you read before you write.
If you want to assign to the to the outer variable from the inner scope you have to cheat like so:
def function1():
al = [10]
def function2():
print al[0]
al[0] -= 1
if al[0]>0:
function2()
function2()
So al is immutable but its contents are not and you can change them without creating a new variable.
It should be noted that this is a syntax glitch in Python. Python itself (at the bytecode level) can assign to these variables just fine; there's simply no syntax in 2.x to indicate that you want to do so. It assumes that if you assign to a variable in a nesting level, you mean for it to be a local to it.
This is a huge shortcoming; being able to assign to closures is fundamental. I've worked around this with charlieb's hack several times.
Python 3 fixes this with the very awkwardly-named "nonlocal" keyword:
def function1():
a = 10
def function2():
nonlocal a
print(a)
a -= 1
if a>0:
function2()
function2()
function1()
It's very poor that this syntax is only available in 3.x; most people are stuck in 2.x, and have to continue using hacks to work around this problem. This badly needs to be backported to 2.x.
http://www.python.org/dev/peps/pep-3104/
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