All variables in Python are references. Elementary data types aren't an exception.

In the first example, you reassign b. It no longer references the same object as a.

In the second example, you modify b. Since you've previously set a and b to be references to the same object, the modification applies to a as well.

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.

Whatever is associated with a variable name has to be stored in the program's memory somewhere. An easy way to think of this, is that every byte of memory has an index-number. For simplicity's sake, lets imagine a simple computer, these index-numbers go from 0 (the first byte), upwards to however many bytes there are.

Say we have a sequence of 37 bytes, that a human might interpret as some words:

"The Owl and the Pussy-cat went to sea"

The computer is storing them in a contiguous block, starting at some index-position in memory. This index-position is most often called an "address". Obviously this address is absolutely just a number, the byte-number of the memory these letters are residing in.

@12000 The Owl and the Pussy-cat went to sea

So at address 12000 is a T, at 12001 an h, 12002 an e ... up to the last a at 12037.

I am labouring the point here because it's fundamental to every programming language. That 12000 is the "address" of this string. It's also a "reference" to it's location. For most intents and purposes an address is a pointer is a reference. Different languages have differing syntactic handling of these, but essentially they're the same thing - dealing with a block of data at a given number.

Python and Java try to hide this addressing as much as possible, where languages like C are quite happy to expose pointers for exactly what they are.

The take-away from this, is that an object reference is the number of where the data is stored in memory. (As is a pointer.)

Now, most programming languages distinguish between simple types: characters and numbers, and complex types: strings, lists and other compound-types. This is where the reference to an object makes a difference.

So when performing operations on simple types, they are independent, they each have their own memory for storage. Imagine the following sequence in python:

>>> a = 3
>>> b = a
>>> b
3
>>> b = 4
>>> b
4
>>> a
3      # <-- original has not changed

The variables a and b do not share the memory where their values are stored. But with a complex type:

>>> s = [ 1, 2, 3 ]
>>> t = s
>>> t
[1, 2, 3]
>>> t[1] = 8
>>> t
[1, 8, 3]
>>> s
[1, 8, 3]  # <-- original HAS changed

We assigned t to be s, but obviously in this case t is s - they share the same memory. Wait, what! Here we have found out that both s and t are a reference to the same object - they simply share (point to) the same address in memory.

One place Python differs from other languages is that it considers strings as a simple type, and these are independent, so they behave like numbers:

>>> j = 'Pussycat'
>>> k = j
>>> k
'Pussycat'
>>> k = 'Owl'
>>> j
'Pussycat'  # <-- Original has not changed

Whereas in C strings are definitely handled as complex types, and would behave like the Python list example.

The upshot of all this, is that when objects that are handled by reference are modified, all references-to this object "see" the change. So if the object is passed to a function that modifies it (i.e.: the content of memory holding the data is changed), the change is reflected outside that function too.

But if a simple type is changed, or passed to a function, it is copied to the function, so the changes are not seen in the original.

For example:

def fnA( my_list ):
    my_list.append( 'A' )

a_list = [ 'B' ]
fnA( a_list )
print( str( a_list ) )
['B', 'A']        # <-- a_list was changed inside the function

But:

def fnB( number ):
    number += 1

x = 3
fnB( x )
print( x )
3                # <-- x was NOT changed inside the function

So keeping in mind that the memory of "objects" that are used by reference is shared by all copies, and memory of simple types is not, it's fairly obvious that the two types operate differently.

In Python a variable is not a box that holds things, it is a name that points to an object. In your code:

• d = { 'm': [1,2,3] } --> binds the name d to a dictionary
• m = d['m'] --> binds the name m to a list
• m = m[1:] --> binds the name m to another list

Your third line is not changing m itself, but what m is pointing to.

To edit the elements in the list what you can do is:

m = d['m']
for i, item in enumerate(m):
    result = do_something_with(item)
    m[i] = result