Python scope problems only when _assigning_ to a variable

Question

So I'm having a very strange error right now. I found where it happens, and here's the simplest code that can reproduce it.

def cause_an_error():
 some_var = False
 def some_nested_func():
 print some_var
 #some_var = True
 some_nested_func()
>>> cause_an_error()
False

If you run it as-is, it prints "False". But if you uncomment that line, it gives an error:

 Traceback (most recent call last):
 File "<pyshell#224>", line 1, in <module>
 cause_an_error()
 File "<pyshell#223>", line 6, in cause_an_error
 some_nested_func()
 File "<pyshell#223>", line 4, in some_nested_func
 print some_var
UnboundLocalError: local variable 'some_var' referenced before assignment

Notice the error occurs on the line that worked just fine before.

Any ideas what causes this and how I can fix this? I'm using Python 2.6.1.

• Ruggero Turra: very similar to this stackoverflow.com/questions/2516652/…

Answer 1

The inner function, some_nest_func, has its own scope, and the fact that you are assigning to some_var somewhere within that scope makes some_var local to that scope. This has the effect of hiding the some_var assigned in the outer function. They are two different variables.

The upshot of all this is that you are printing some_nest_func's version of some_var before you assign to it.

If you need to manipulate some variable both in and outside the inner function, there isn't a clean way to it, AFAIK. A simple work-around is to replace the variable with a single-element array and manipulate that isn't (don't assign to the array variable, of course).

• Ponkadoodle: That makes sense to me. But shouldn't the non-local variable only be shadowed after the assignment?
• Marcelo Cantos: The shadowing happens at compile time. If Python finds an assignment anywhere within the scope, all references become local.
• Marcelo Cantos: I've amended my answer. Also see Ignacio's answer for Python 3.0.

Answer 2

This can't be fixed in 2.x without refactoring the code. 3.x adds nonlocal to solve this.

Answer 3

You can do this by changing a value of a referenced (read mutable) variable such as dict, although ugly, this solution (transcribed from an answer to a similar question) should work:

def avoid_an_error():
 D = {"some_var": False}
 def some_nested_func():
 D["some_var"] = True
 print D["some_var"]
 some_nested_func()
>>> avoid_an_error()
True