local variable 'a' referenced b

[Dave Angel]

On 10/02/2012 10:03 PM, contro opinion wrote:
> code1
>>>> def foo():
> ...     a = 1
> ...     def bar():
> ...         b=2
> ...         print a + b
> ...     bar()
> ...
> ...
>>>> foo()
> 3
>
> code2
>>>> def foo():
> ...     a = 1
> ...     def bar():
> ...         b=2
> ...         a = a + b

Because your function bar() has an assignment to a, it becomes a local,
and masks access to the one in the containing function.

Then because when you start executing that assignment statement, a
hasn't yet gotten a value, you get the error below.

> ...         print a
> ...     bar()
> ...
>>>> foo()
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
>   File "<stdin>", line 7, in foo
>   File "<stdin>", line 5, in bar
> UnboundLocalError: local variable 'a' referenced b
>
> why code2 can not get output of 3?
>

In Python3, you can avoid the "problem" by declaring a as nonlocal.


def foo():
     a = 1
     def bar():
         nonlocal a
         b=2
         a = a + b
         print (a)
     bar()

foo()

if you're stuck with Python2.x, you can use a mutable object for a, and
mutate it, rather than replace it.  For example,


def foo():
     a = [3]
     def bar():
         b=2
         a.append(b)   #this mutates a, but doesn't assign it
         print (a)
         a[0] += b  #likewise, for a number within the list
         print (a)
     bar()

That should work in either 2.x or 3.2

-- 

DaveA