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

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.

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 = []
    ...

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().

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. 🙂

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.

Contrast 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.

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.

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)

Please allow me to edit again. These concepts are my experience from learning python by try error and internet, mostly stackoverflow. There are mistakes and there are helps.

Python variables use references, I think reference as relation links from name, memory adress and value.

When we do B = A, we actually create a nickname of A, and now the A has 2 names, A and B. When we call B, we actually are calling the A. we create a ink to the value of other variable, instead of create a new same value, this is what we call reference. And this thought would lead to 2 porblems.

when we do

A = [1]
B = A   # Now B is an alias of A

A.append(2)  # Now the value of A had been changes
print(B)
>>> [1, 2]  
# B is still an alias of A
# Which means when we call B, the real name we are calling is A

# When we do something to B,  the real name of our object is A
B.append(3)
print(A)
>>> [1, 2, 3]

This is what happens when we pass arguments to functions

def test(B):
    print('My name is B')
    print(f'My value is {B}') 
    print(' I am just a nickname,  My real name is A')
    B.append(2)


A = [1]
test(A) 
print(A)
>>> [1, 2]

We pass A as an argument of a function, but the name of this argument in that function is B.
Same one with different names.
So when we do B.append, we are doing A.append
When we pass an argument to a function, we are not passing a variable , we are passing an alias.

And here comes the 2 problems.

1. the equal sign always creates a new name

A = [1]
B = A
B.append(2)
A = A[0]  # Now the A is a brand new name, and has nothing todo with the old A from now on.

B.append(3)
print(A)
>>> 1
# the relation of A and B is removed when we assign the name A to something else
# Now B is a independent variable of hisown.

the Equal sign is a statesment of clear brand new name,

this was the concused part of mine

 A = [1, 2, 3]

# No equal sign, we are working on the origial object,
A.append(4)
>>> [1, 2, 3, 4]

# This would create a new A
A = A + [4]  
>>> [1, 2, 3, 4]

and the function

def test(B):
    B = [1, 2, 3]   # B is a new name now, not an alias of A anymore
    B.append(4)  # so this operation won't effect A
    
A = [1, 2, 3]
test(A)
print(A)
>>> [1, 2, 3]

# ---------------------------

def test(B):
    B.append(4)  # B is a nickname of A, we are doing A
    
A = [1, 2, 3]
test(A)
print(A)
>>> [1, 2, 3, 4]

the first problem is

1. the left side of and equation is always a brand new name, new variable,

2. unless the right side is a name, like B = A, this create an alias only

The second problem, there are something would never be changed, we cannot modify the original, can only create a new one.

This is what we call immutable.

When we do A= 123 , we create a dict which contains name, value, and adress.

When we do B = A, we copy the adress and value from A to B, all operation to B effect the same adress of the value of A.

When it comes to string, numbers, and tuple. the pair of value and adress could never be change. When we put a str to some adress, it was locked right away, the result of all modifications would be put into other adress.

A = 'string' would create a protected value and adess to storage the string ‘string’ . Currently, there is no built-in functions or method cound modify a string with the syntax like list.append, because this code modify the original value of a adress.

the value and adress of a string, a number, or a tuple is protected, locked, immutable.

All we can work on a string is by the syntax of A = B.method , we have to create a new name to storage the new string value.

please extend this discussion if you still get confused.
this discussion help me to figure out mutable / immutable / refetence / argument / variable / name once for all, hopely this could do some help to someone too.

##############################

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.

As jouell said. It’s a matter of what points to what and i’d add that it’s also a matter of the difference between what = does and what the .append method does.

1. When you define n and x in main, you tell them to point at 2 objects, namely 1 and [1,2,3]. That is what = does : it tells what your variable should point to.

2. When you call the function f(n,x), you tell two new local variables nf and xf to point at the same two objects as n and x.

3. When you use "something"="anything_new", you change what "something" points to. When you use .append, you change the object itself.

4. Somehow, even though you gave them the same names, n in the main() and the n in f() are not the same entity, they only originally point to the same object (same goes for x actually). A change to what one of them points to won’t affect the other. However, if you instead make a change to the object itself, that will affect both variables as they both point to this same, now modified, object.

Lets illustrate the difference between the method .append and the = without defining a new function :

compare

    m = [1,2,3]
    n = m   # this tells n to point at the same object as m does at the moment
    m = [1,2,3,4] # writing m = m + [4] would also do the same
    print('n = ', n,'m = ',m)

to

    m = [1,2,3]
    n = m
    m.append(4)
    print('n = ', n,'m = ',m)

In the first code, it will print n = [1, 2, 3] m = [1, 2, 3, 4], since in the 3rd line, you didnt change the object [1,2,3], but rather you told m to point to a new, different, object (using ‘=’), while n still pointed at the original object.

In the second code, it will print n = [1, 2, 3, 4] m = [1, 2, 3, 4]. This is because here both m and n still point to the same object throughout the code, but you modified the object itself (that m is pointing to) using the .append method… Note that the result of the second code will be the same regardless of wether you write m.append(4) or n.append(4) on the 3rd line.

Once you understand that, the only confusion that remains is really to understand that, as I said, the n and x inside your f() function and the ones in your main() are NOT the same, they only initially point to the same object when you call f().

When you are passing the command n = 2 inside the function, it finds a memory space and label it as 2. But if you call the method append, you are basically refrencing to location x (whatever the value is) and do some operation on that.