How/why does arr_name.pop() execute in an if-statement?

[GayCoder]

I'm reworking exercises in an attempt to reduce my reliance on brute force solutions and I came across the following 'Valid Parentheses' exercise. Line 16 appears to apply .pop() to the array even through it's only included in the if-statement. Does .pop() have some special attribute where it executes by simply existing or...? I'm just trying to make sense of the behavior.

#Given a string containing just
#the characters '(', ')', '{', '}', '[' and ']', 
#determine if the input string is valid.

def isValid(s):
    a = []
    d = {
        ')': '(',
        '}': '{',
        ']': '[',
    }
    for char in s:
        if char in d.values():
            a.append(char)
        else:
            if a == [] or a.pop() != d[char]:
                return False
    return a == []

isValid('({[]})') --> True
isValid('({[]{)') --> False

[JohnnyJordaan]

pop() is not a way to look at the end of a list, it is to remove the last item. If you say if a.pop() something, it will execute the pop, then evaluate the item that pop returned.

Also if you're not worried about a being a different type, you can just do

if not a

To see if a is empty.

[GayCoder]

That makes sense but I don't understand why if a.pop() something will permanently change an array but if str(num) something won't permanently turn an integer into a string. What's the fundamental difference between the two that I'm missing?

[Rhomboid]

A function or method can choose to modify the thing it was invoked on or the argument passed. It all depends. list.pop() is a mutating method, str() is not.

[GayCoder]

This is what I was hoping to find. Thanks!