Free variables treated as globals in Python?

Question

In the Execution Model section of the Python 3.7 reference manual I read the following statement:

The global statement has the same scope as a name binding operation in the same block. If the nearest enclosing scope for a free variable contains a global statement, the free variable is treated as a global.

So I typed the following code into the Python Interpreter:

x =0
def func1():
    global x
    def func2():
        x = 1
    func2()

After calling func1() I would have expected the value of x in the global scope to change to 1.

What did I get wrong?

Martijn Pieters · Accepted Answer · 2018-10-05T15:12:27.830

x = 1 in func2 is not a free variable. It's just another local; you are binding to the name and names bound to are, by default, locals unless you tell Python otherwise.

From the same Execution model documentation:

If a name is bound in a block, it is a local variable of that block, unless declared as nonlocal or global. [...] If a variable is used in a code block but not defined there, it is a free variable.

(Bold emphasis mine)

You bound the name in the block with x = 1, so it is a local variable in that block, and can't be a free variable. So section you found doesn't apply, because that would only apply to free variables:

If the nearest enclosing scope for a free variable contains a global statement, the free variable is treated as a global.

You should not bind x in func2(), because only names that are not binding in the scope are free variables.

So this works:

>>> def func1():
...     global x
...     x = 1
...     def func2():
...         print(x)  # x is a free variable here
...     func2()
...
>>> func1()
1
>>> x
1

x in func2 is now a free variable; it is not defined in the scope of func2, so picks it up from the parent scope. The parent scope here is func1, but x is marked a global there so when reading x for the print() function the global value is used.

Contrast this with x not being marked as a global in func1:

>>> def func1():
...     x = 1
...     def func2():
...         print(x)  # x is free variable here, now referring to x in func1
...     func2()
...
>>> x = 42
>>> func1()
1

Here the global name x is set to 42, but this doesn't affect what is printed. x in func2 is a free variable, but the parent scope func1 only has x as a local name.

It becomes all the more interesting when you add a new outer-most scope where x is still local:

>>> def outerfunc():
...     x = 0   # x is a local
...     def func1():
...         global x   # x is global in this scope and onwards
...         def func2():
...             print('func2:', x)  # x is a free variable
...         func2()
...     print('outerfunc:', x)
...     func1()
...
>>> x = 42
>>> outerfunc()
outerfunc: 0
func2: 42
>>> x = 81
>>> outerfunc()
outerfunc: 0
func2: 81

x in outerfunc is bound, so not a free variable. It is therefore a local in that scope. However, in func1, the global x declaration marks x as a global in the nested scrope. In func2 x is a free variable, and by the statement that you found, it is treated as a global.

Free variables treated as globals in Python?

1 Answers1