Python Referencing – How Python Referencing Works

Arguments are passed by assignment. The rationale behind this is twofold:

1. the parameter passed in is actually a reference to an object (but the reference is passed by value)
2. some data types are mutable, but others aren't

So:

• If you pass a mutable object into a method, the method gets a reference to that same object and you can mutate it to your heart's delight, but if you rebind the reference in the method, the outer scope will know nothing about it, and after you're done, the outer reference will still point at the original object.

• If you pass an immutable object to a method, you still can't rebind the outer reference, and you can't even mutate the object.

To make it even more clear, let's have some examples.

List - a mutable type

Let's try to modify the list that was passed to a method:

def try_to_change_list_contents(the_list):
    print('got', the_list)
    the_list.append('four')
    print('changed to', the_list)

outer_list = ['one', 'two', 'three']

print('before, outer_list =', outer_list)
try_to_change_list_contents(outer_list)
print('after, outer_list =', outer_list)

Output:

before, outer_list = ['one', 'two', 'three']
got ['one', 'two', 'three']
changed to ['one', 'two', 'three', 'four']
after, outer_list = ['one', 'two', 'three', 'four']

Since the parameter passed in is a reference to outer_list, not a copy of it, we can use the mutating list methods to change it and have the changes reflected in the outer scope.

Now let's see what happens when we try to change the reference that was passed in as a parameter:

def try_to_change_list_reference(the_list):
    print('got', the_list)
    the_list = ['and', 'we', 'can', 'not', 'lie']
    print('set to', the_list)

outer_list = ['we', 'like', 'proper', 'English']

print('before, outer_list =', outer_list)
try_to_change_list_reference(outer_list)
print('after, outer_list =', outer_list)

Output:

before, outer_list = ['we', 'like', 'proper', 'English']
got ['we', 'like', 'proper', 'English']
set to ['and', 'we', 'can', 'not', 'lie']
after, outer_list = ['we', 'like', 'proper', 'English']

Since the the_list parameter was passed by value, assigning a new list to it had no effect that the code outside the method could see. The the_list was a copy of the outer_list reference, and we had the_list point to a new list, but there was no way to change where outer_list pointed.

String - an immutable type

It's immutable, so there's nothing we can do to change the contents of the string

Now, let's try to change the reference

def try_to_change_string_reference(the_string):
    print('got', the_string)
    the_string = 'In a kingdom by the sea'
    print('set to', the_string)

outer_string = 'It was many and many a year ago'

print('before, outer_string =', outer_string)
try_to_change_string_reference(outer_string)
print('after, outer_string =', outer_string)

Output:

before, outer_string = It was many and many a year ago
got It was many and many a year ago
set to In a kingdom by the sea
after, outer_string = It was many and many a year ago

Again, since the the_string parameter was passed by value, assigning a new string to it had no effect that the code outside the method could see. The the_string was a copy of the outer_string reference, and we had the_string point to a new string, but there was no way to change where outer_string pointed.

I hope this clears things up a little.

EDIT: It's been noted that this doesn't answer the question that @David originally asked, "Is there something I can do to pass the variable by actual reference?". Let's work on that.

How do we get around this?

As @Andrea's answer shows, you could return the new value. This doesn't change the way things are passed in, but does let you get the information you want back out:

def return_a_whole_new_string(the_string):
    new_string = something_to_do_with_the_old_string(the_string)
    return new_string

# then you could call it like
my_string = return_a_whole_new_string(my_string)

If you really wanted to avoid using a return value, you could create a class to hold your value and pass it into the function or use an existing class, like a list:

def use_a_wrapper_to_simulate_pass_by_reference(stuff_to_change):
    new_string = something_to_do_with_the_old_string(stuff_to_change[0])
    stuff_to_change[0] = new_string

# then you could call it like
wrapper = [my_string]
use_a_wrapper_to_simulate_pass_by_reference(wrapper)

do_something_with(wrapper[0])

Although this seems a little cumbersome.

I can see three fundamental Python concepts that can shine some light on the question:

1) First, an assignment from a mutable object like in

self.foo = arg1

is like copying a pointer (and not the value pointed to): self.foo and arg1 are the same object. That's why the line that follows,

self.foo.age = 23

modifies arg1 (i.e. Person1). Variables are thus different "names" or "labels" that can point to a unique object (here, a person object). This explains why baz(Person1) modifies Person1.age and bar1.foo.age to 27, since Person1 and bar1.foo are just two names for the same person object (Person1 is bar1.foo returns True, in Python).

2) The second important notion is that of assignments. In

def teh(arg1):
    arg1 = [3,2,1]

variable arg1 is local, so that the code

meh = [1,2,3]
teh(meh)

first does arg1 = meh, which means that arg1 is an additional (local) name for list meh; but doing arg1 = [3, 2, 1] is like saying "I changed my mind: arg1 will from now on be the name of a new list, [3, 2, 1]". The important thing to keep in mind here is that assignments, despite being denoted with an "equal" sign, are asymmetrical: they give to a (mutable) object on the right-and-side an additional name, given in the left-hand side (that's why you can't do [3, 2, 1] = arg1, as the left-hand side must be a name [or names]). So, arg1 = meh; arg1 = [3, 2, 1] cannot change meh.

3) The last point is related to the question title: "passing by value" and "passing by reference" are not concepts that are relevant in Python. The relevant concepts are instead "mutable object" and "immutable object". Lists are mutable, while numbers are not, which explains what you observe. Also, your Person1 and bar1 objects are mutable (that's why you can change the person's age). You can find more information about these notions in a text tutorial and a video tutorial. Wikipedia also has some (more technical) information. An example illustrates the difference of behavior between mutable and immutable:

x = (4, 2)
y = x  # This is equivalent to copying the value (4, 2), because tuples are immutable
y += (1, 2, 3)  # This does not change x, because tuples are immutable; this does y = (4, 2) + (1, 2, 3)

x = [4, 2]
y = x  # This is equivalent to copying a pointer to the [4, 2] list
y += [1, 2, 3]  # This also changes x, because x and y are different names for the same (mutable) object

The last line is not equivalent to y = y + [1, 2, 3] because this would only put a new list object in variable y instead of changing the list referred to by both y and x.

The three concepts above (variables as names [for mutable objects], asymmetrical assignment, and mutability/immutability) explain many of Python's behaviors.