Python revise global variables inside function

Question

Background

Suppose I have 2 global variables a = [1,2,3] and b = ['a','b'], If I must exchange a and b inside a function, I can do with following code:

a = [1,2,3]
b = ['a','b']
def fun1():
    global a,b
    temp = a[::-1]
    a = b[::-1]
    b = temp
fun1()
print(a,b)

Out:

['b', 'a'] [3, 2, 1]

However, If a and b are parameters of my function, how do we do it?

a = [1,2,3]
b = ['a','b']
def fun2(a,b):
    global a,b
    temp = a[::-1]
    a = b[::-1]
    b = temp
fun2(a,b)
print(a,b)

Raise Error:

SyntaxError: name 'a' is parameter and global

My attempt

I tried to do with exec, it works on Python 3.7, However, it does not work on Codewars 3.6 env. I don't know the reason, maybe I just find wrong method.

a = [1,2,3]
b = ['a','b']
def exchange_with(a, b):
    temp = a[::-1].copy()
    exec('a = b[::-1]',globals())
    exec('b = temp',locals(),globals())
exchange_with(a, b)
a,b

Out:

(['b', 'a'], [3, 2, 1])

You can't have a variable as an parameter and global you have to choose. It's better not to use global, just delete it. — ErnestBidouille
– ErnestBidouille, Commented Feb 3, 2020 at 16:26

chepner · Accepted Answer · 2020-02-03 16:20:11Z

1

You still have to pass the values as arguments. Once inside the function, you need to make in-place slice assignments to each list. No global names need be involved.

def fun2(a,b):
    a[:], b[:] = b[::-1], a[::-1]

x = [1,2,3]
y = ['a', 'b']
fun2(x, y)
assert x == ['b', 'a']
assert y == [3, 2, 1]

answered Feb 3, 2020 at 16:20

chepner

538k77 gold badges594 silver badges746 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

Travis Over a year ago

I used to treat a and b inside function body as local variable, which means a inside function body only contains value of global a. However, your method is absolutely right! id of global a == id(a) inside function. I think it is so dangerous that we can revise global variable a without a global a statement. So in practice, better to not change global variables like this?

chepner Over a year ago

a and b are local variables, but that doesn't mean they can't refer to values that are also referred to by global variables, which means that if you aren't careful, you can mutate global state using a local variables. (If you haven't already, you may want to read nedbatchelder.com/text/names.html.) In other words, names have scope, but values do not.

Bowen Liu Over a year ago

@chepner Not OP but thanks for your reply. However, the link you posted doesn't work, care to give it an update? I followed here from your CodeWars answer. My question is: why [:] will make it work when the function is a, b = b[::-1], a[::-1], the global values don't get changed at all? Thanks.

chepner Over a year ago

Sorry, the link should be nedbatchelder.com/text/names.html; the end-of-sentence period got included.

chepner Over a year ago

a = ... is an ordinary assignment; it makes the name a refer to a new value regardless of what it used to refer to. a[:] = ... isn't "really" an assignment at all; it's implemented by a method call a.__setitem__(slice(None, None, None), ...).

|

Collectives™ on Stack Overflow

Python revise global variables inside function

Background

My attempt

1 Answer 1

7 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Background

My attempt

1 Answer 1

7 Comments

Your Answer

Sign up or log in

Post as a guest

Related