[python] Why can a function modify some arguments as perceived by the caller, but not others?

I'm trying to understand Python's approach to variable scope. In this example, why is f() able to alter the value of x, as perceived within main(), but not the value of n?

def f(n, x):
    n = 2
    x.append(4)
    print('In f():', n, x)

def main():
    n = 1
    x = [0,1,2,3]
    print('Before:', n, x)
    f(n, x)
    print('After: ', n, x)

main()

Output:

Before: 1 [0, 1, 2, 3]
In f(): 2 [0, 1, 2, 3, 4]
After:  1 [0, 1, 2, 3, 4]

I will rename variables to reduce confusion. n -> nf or nmain. x -> xf or xmain:

def f(nf, xf):
    nf = 2
    xf.append(4)
    print 'In f():', nf, xf

def main():
    nmain = 1
    xmain = [0,1,2,3]
    print 'Before:', nmain, xmain
    f(nmain, xmain)
    print 'After: ', nmain, xmain

main()

When you call the function f, the Python runtime makes a copy of xmain and assigns it to xf, and similarly assigns a copy of nmain to nf.

In the case of n, the value that is copied is 1.

In the case of x the value that is copied is not the literal list [0, 1, 2, 3]. It is a reference to that list. xf and xmain are pointing at the same list, so when you modify xf you are also modifying xmain.

If, however, you were to write something like:

    xf = ["foo", "bar"]
    xf.append(4)

you would find that xmain has not changed. This is because, in the line xf = ["foo", "bar"] you have change xf to point to a new list. Any changes you make to this new list will have no effects on the list that xmain still points to.

Hope that helps. :-)

n is an int (immutable), and a copy is passed to the function, so in the function you are changing the copy.

X is a list (mutable), and a copy of the pointer is passed o the function so x.append(4) changes the contents of the list. However, you you said x = [0,1,2,3,4] in your function, you would not change the contents of x in main().

You've got a number of answers already, and I broadly agree with J.F. Sebastian, but you might find this useful as a shortcut:

Any time you see varname =, you're creating a new name binding within the function's scope. Whatever value varname was bound to before is lost within this scope.

Any time you see varname.foo() you're calling a method on varname. The method may alter varname (e.g. list.append). varname (or, rather, the object that varname names) may exist in more than one scope, and since it's the same object, any changes will be visible in all scopes.

[note that the global keyword creates an exception to the first case]

Python is a pure pass-by-value language if you think about it the right way. A python variable stores the location of an object in memory. The Python variable does not store the object itself. When you pass a variable to a function, you are passing a copy of the address of the object being pointed to by the variable.

Contrasst these two functions

def foo(x):
    x[0] = 5

def goo(x):
    x = []

Now, when you type into the shell

>>> cow = [3,4,5]
>>> foo(cow)
>>> cow
[5,4,5]

Compare this to goo.

>>> cow = [3,4,5]
>>> goo(cow)
>>> goo
[3,4,5]

In the first case, we pass a copy the address of cow to foo and foo modified the state of the object residing there. The object gets modified.

In the second case you pass a copy of the address of cow to goo. Then goo proceeds to change that copy. Effect: none.

I call this the pink house principle. If you make a copy of your address and tell a painter to paint the house at that address pink, you will wind up with a pink house. If you give the painter a copy of your address and tell him to change it to a new address, the address of your house does not change.

The explanation eliminates a lot of confusion. Python passes the addresses variables store by value.

My general understanding is that any object variable (such as a list or a dict, among others) can be modified through its functions. What I believe you are not able to do is reassign the parameter - i.e., assign it by reference within a callable function.

That is consistent with many other languages.

Run the following short script to see how it works:

def func1(x, l1):
    x = 5
    l1.append("nonsense")

y = 10
list1 = ["meaning"]
func1(y, list1)
print(y)
print(list1)

It´s because a list is a mutable object. You´re not setting x to the value of [0,1,2,3], you´re defining a label to the object [0,1,2,3].

You should declare your function f() like this:

def f(n, x=None):
    if x is None:
        x = []
    ...

Python is copy by value of reference. An object occupies a field in memory, and a reference is associated with that object, but itself occupies a field in memory. And name/value is associated with a reference. In python function, it always copy the value of the reference, so in your code, n is copied to be a new name, when you assign that, it has a new space in caller stack. But for the list, the name also got copied, but it refer to the same memory(since you never assign the list a new value). That is a magic in python!

If the functions are re-written with completely different variables and we call id on them, it then illustrates the point well. I didn't get this at first and read jfs' post with the great explanation, so I tried to understand/convince myself:

def f(y, z):
    y = 2
    z.append(4)
    print ('In f():             ', id(y), id(z))

def main():
    n = 1
    x = [0,1,2,3]
    print ('Before in main:', n, x,id(n),id(x))
    f(n, x)
    print ('After in main:', n, x,id(n),id(x))

main()
Before in main: 1 [0, 1, 2, 3]   94635800628352 139808499830024
In f():                          94635800628384 139808499830024
After in main: 1 [0, 1, 2, 3, 4] 94635800628352 139808499830024

z and x have the same id. Just different tags for the same underlying structure as the article says.

f doesn't actually alter the value of x (which is always the same reference to an instance of a list). Rather, it alters the contents of this list.

In both cases, a copy of a reference is passed to the function. Inside the function,

• n gets assigned a new value. Only the reference inside the function is modified, not the one outside it.
• x does not get assigned a new value: neither the reference inside nor outside the function are modified. Instead, x’s value is modified.

Since both the x inside the function and outside it refer to the same value, both see the modification. By contrast, the n inside the function and outside it refer to different values after n was reassigned inside the function.

I had modified my answer tons of times and realized i don't have to say anything, python had explained itself already.

a = 'string'
a.replace('t', '_')
print(a)
>>> 'string'

a = a.replace('t', '_')
print(a)
>>> 's_ring'

b = 100
b + 1
print(b)
>>> 100

b = b + 1
print(b)
>>> 101

def test_id(arg):
    c = id(arg)
    arg = 123
    d = id(arg)
    return

a = 'test ids'
b = id(a)
test_id(a)
e = id(a)

# b = c  = e != d

# this function do change original value
del change_like_mutable(arg):
    arg.append(1)
    arg.insert(0, 9)
    arg.remove(2)
    return

test_1 = [1, 2, 3]
change_like_mutable(test_1)



# this function doesn't 
def wont_change_like_str(arg):
     arg = [1, 2, 3]
     return


test_2 = [1, 1, 1]
wont_change_like_str(test_2)
print("Doesn't change like a imutable", test_2)

This devil is not the reference / value / mutable or not / instance, name space or variable / list or str, IT IS THE SYNTAX, EQUAL SIGN.