分配与全局变量同名的局部变量时出错

Question

I am seeing an error when assigning either a global or an enclosing-function local to a local variable with the same name. The code below illustrates this issue, where f() runs fine, while g() raises an error. It seems like python knows that a is being assigned locally, so it says that all references to a are now local, even the references before a is actually assigned locally. What explains this behavior? I am running Python 2.7.12 :: Anaconda 4.2.0 (64-bit).

In [18]: a = 1
    ...: 
    ...: def f():
    ...:   x = a
    ...:   print x
    ...: 
    ...: def g():
    ...:   a = a
    ...:   print a
    ...:   

In [19]: f()
1

In [20]: g()
---------------------------------------------------------------------------
UnboundLocalError                         Traceback (most recent call last)
<ipython-input-20-d65ffd94a45c> in <module>()
----> 1 g()

<ipython-input-18-f3d970bdaa2b> in g()
      6 
      7 def g():
----> 8   a = a
      9   print a
     10 

UnboundLocalError: local variable 'a' referenced before assignment

Answer 1

The short answer is that, within g(), you need to declare

global a

if you want to be able to modify "a" from within a function and have this effect globally visible. However, in your case, the effect of using "a" within g() is to convert this variable name to refer to a local-scope variable, which then hides the global "a" which you're attempting to use on the righthand side of your assignment, triggering the exception. This is more fully explained here . There is also a Python FAQ which explains the rules that make f() work without the need for the "global a".