Function changes the global list in python instead of local list

Question

I have this code written in Python:

def trick(l,i):
    if i==1:
        return l[0];
    else :
        return l[1]
def foo(x):
    x[0] = 'def'
    x[1] = 'abc'
    return x[1]
q = ['abc', 'def']
print(trick(q,1) == foo(q))
print(trick(q,1) == foo(q))
print(trick(q,0) == foo(q))

The output is this:

True
False
True

Why does the second print statement print "False" even though the second and the first print statements are the same. And when I visualized this code in Python tutor I got to know that the foo(x) function actually changes/swaps the elements in the original/global list 'q' even though 'x' should have local scope.

I can't understand this. Please help.

x is mutable. Nothing to do with the scope. you changed the items in x — Jean-François Fabre
– Jean-François Fabre ♦, Commented Oct 23, 2017 at 19:43
You don't have separate global and local lists. There's only one list here. See nedbatchelder.com/text/names.html — user2357112
– user2357112, Commented Oct 23, 2017 at 19:43
@Jean-FrançoisFabre Yep, I changed the items in x, not in q. so in the second print statement, I am calling the same statement but the output is False. Why so? I thought q remained the same and only x changed. — Parker Queen
– Parker Queen, Commented Oct 23, 2017 at 19:44
@user2357112 Thank you so much. You can reply as an answer so I can mark it as the correct one. — Parker Queen
– Parker Queen, Commented Oct 23, 2017 at 19:49

user2357112 · Accepted Answer · 2017-10-23 19:55:34Z

1

While you do have separate global and local variables, you don't have separate global and local lists. Both q and x refer to the same list; the parameter passing mechanism doesn't make an implicit copy of the list. (Python almost never makes implicit copies.)

answered Oct 23, 2017 at 19:55

user2357112

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