为什么全局关键字即使无法访问也不会被忽略？ （Python）

Question

I am really new to programming and Python , so please really forgive me for my ignorance.

I just learned that using global can make a variable inside function be a global one. However, I discovered something not with my expectation:

I tried the following code in Python 3.8 (forgive me for my ignorance as I don't know what else information I should provide):

>>> x = 0
>>> 
>>> def function():
...     if False:
...             global x
...     x = 1
...
>>> function()
>>> print(x)

and the result is 1.

However, I expected the code to have the same effect as the following code:

>>> x = 0
>>> 
>>> def function():
...     x = 1
...
>>> function()
>>> print(x)

which the result should be 0.

In my mind, the statement inside if False should not be executed, so it sounds strange to me. Also, personally, I think that in some situation I would expect the variable inside a function, whether local or global, to be dependent on other codes... what I mean is, I would like to change if False to something like if A == 'A' , while (I hope) I can control whether the x is global/local according to my conditional statement.

I tried to change if to while , but it's the same... there isn't a infinite loop, but the code global x is still executed/compiled...

I admit that it may sounds naive, and perhaps it just won't work in Python , but I really wonder why... It seems that the code global x is unreachable, but how come it is not ignored?

Can anyone please tell me about the reason? I would like to know more about the mechanism behind compilation(?)

Any help would be appreciated, thank you!

Answer 1

In python the global statement (and the nonlocal statement) are very different from the normal python code. Essentially no matter where a global statement in a function is, it influences always the current codeblock, and is never "executed". You should think more of it as a compiler directive instead of a command.

Note that the statement itself must come before any usage of the variable it modifies, ie

print(x)
global x

is a syntax error. The global statement can only modify variable behavior in the whole codeblock, you can't first have a non-global variable that later gets global and you can also not have conditional global variable

(I couldn't really find good documentation for this behavior, here it says "The global statement is a declaration which holds for the entire current code block." and "global is a directive to the parser. It applies only to code parsed at the same time as the global statement." but that doesn't seem super clear to me.)

There are more compiler directives in python, although they don't always look like one. One is the from __future__ import statements which look like module imports but change python behavior.

Answer 2

Global is not in execution path but in a scope. The scope is whole function. Statements like if for don't make scopes. If you use any assignment you create local variable. The same with global or nonlocal you bind symbol to variable from outside.

As Stanislas Morbieu typed, see doc .

Programmer's note: global is a directive to the parser. It applies only to code parsed at the same time as the global statement.

Not at execution time.

x = 1
def fun():
    y = x + 1
    print(f'local x = {x}, y = {y}')
fun()
print(f'global x = {x}')

# Output:
#  local x = 1, y = 2
#  global x = 1

In example above, y uses global x (and adds 1 ).

x = 1
def fun():
    y = x
    x = y + 1
    print(f'local x = {x}')
fun()
print(f'global x = {x}')

# Output:
#  UnboundLocalError: local variable 'x' referenced before assignment

Look at last example. It doesn't assign y from global x because assignment in second line creates local x and y can not read local x before x assignment. The same with:

x = 1
def fun():
    if False:
        x += 1
fun()
# Output
#  UnboundLocalError: local variable 'x' referenced before assignment

x assignment creates local variable.

If you want to change global variable under condition you can use globals() .

x = 1
def set_x(do_set, value):
    if do_set:
        globals()['x'] = value
print(f'global x = {x} (init)')
set_x(False, 2)
print(f'global x = {x} (false)')
set_x(True, 3)
print(f'global x = {x} (true)')
# Output
#  global x = 1 (init)
#  global x = 1 (false)
#  global x = 3 (true)

Proxy

I you want to decide with variable you want to use later (in the same scope) you need some kind of proxy IMO.

x = 1
def fun(use_global):
    x = 2  # local
    scope = globals() if use_global else locals()
    scope['x'] += 1
    print(f'local ({use_global}): x = {scope["x"]}')
print(f'global: x = {x} (init)')
fun(False)
print(f'global: x = {x} (false)')
fun(True)
print(f'global: x = {x} (true)')
# Output:
#  global: x = 1 (init)
#  local (False): x = 3
#  global: x = 1 (false)
#  local (True): x = 2
#  global: x = 2 (true)

Maybe you can think about refactoring of your code if you need it.

If you can change local variable name (if not use globals() as above), you can proxy:

• use dict (like in example above)
• use list ( x=[1] ) and usage x[0]
• use object (with builtin dict), example:

class X:
    def __init__(self, x):
        self.x = x

x = X(1)
def fun(use_global):
    global x
    my_x = x if use_global else X(2)
    my_x.x += 1
    print(f'local ({use_global}): x = {my_x.x}')

print(f'global: x = {x.x} (init)')
fun(False)
print(f'global: x = {x.x} (false)')
fun(True)
print(f'global: x = {x.x} (true)')

# Output:
#  global: x = 1 (init)
#  local (False): x = 3
#  global: x = 1 (false)
#  local (True): x = 2
#  global: x = 2 (true)

Note. Variables in Python are only references. It is way you can not change x = 1 without global (or globals() ). You change reference to local value 1 .
But you can change z[0] or z['x'] or zx . Because z referents to list or dict or object and you modify it content.

See: https://realpython.com/python-variables/#object-references
You can check real object by id() function, ex. print(id(x), id(my_x)) .

Answer 3

As per the Python documentation , global is a directive to the parser so it is taken into account before the execution, therefore it does not matter if the code is reachable or not. The variable is global for the entire scope, which is the function in your case.