自由变量在Python中被视为全局变量吗？

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?

Answer 1

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.