Python: Modify Global list inside a function

Question

First of all, I understand that I can use global statement to access global variables. But somehow I was able to modify a global list without global

Answer 1

In this specific case it is because lists are mutable.

As a result having them in the global namespace, or even passed through a function, means that they will be changed as Python holds the reference to the mutable object, not a copy of it.

If you try the same thing with tuples it will not work, as they are immutable.

The way to avoid this is to provide a copy of the list to the function, not the list itself:

func2(list[:])

At the same time you can do this with default arguments, where you can specify a default argument to be [], and if you then .append() something to it, that default argument will forever hold that item within it for all future calls (unless you remove it in some way).

Answer 2

I understand that I can use global statement to access global variables

Your understanding is wrong. You can always access a global variable as long as you don't have a local variable of the same name. You only need the global statement when you are going to change what object a variable name refers to. In your func2, you are not doing that; you are only changing the contents of the object. nums still refers to the same list.

Answer 3

It is of concept based on mutable and immutable objects in Python. In your case, for example:

a=[1,2]
def myfn():
    a=[3,4]
    print id(a)

>>>id(a)
3065250924L
>>>myfn()
3055359596

It is clear both are different objects. Now:

a=[1,2]
def myfn():
    a[:] =[3,4]
    print id(a)

>>>id(a)
3055358572
>>>myfn()
3055358572

That means it is same variable using in local and global scope.

Answer 4

• 2 variables nums are different and they point to a same object or 2 different objects, though they have same name.
• when you call func1(nums), mean that you pass a reference. Now the 2 variable nums point to same object. (2 variables, 1 object)
• when you assign in func1, the inside variable nums will point to a new object, the outside is still unchanged (2 variables, 2 object)
• and when you call print nums then this nums is the outside variable,

Answer 5

There are two reasons for this result:

1. Variables are simply names that refer to objects
2. List is mutable

In func1,nums refer to a new object because a new list is created. Therefore global nums is not affected.

In func2, the modification is applied to the object passed in. Thus global nums is changed. A new object is not created because list is mutable.

ref: https://docs.python.org/3/faq/programming.html#what-are-the-rules-for-local-and-global-variables-in-python