Python Functions Call by Reference

Python functions call by reference

You can not change an immutable object, like str or tuple, inside a function in Python, but you can do things like:

def foo(y):
  y[0] = y[0]**2

x = [5]
foo(x)
print x[0]  # prints 25

That is a weird way to go about it, however, unless you need to always square certain elements in an array.

Note that in Python, you can also return more than one value, making some of the use cases for pass by reference less important:

def foo(x, y):
   return x**2, y**2

a = 2
b = 3
a, b = foo(a, b)  # a == 4; b == 9

When you return values like that, they are being returned as a Tuple which is in turn unpacked.

edit:
Another way to think about this is that, while you can't explicitly pass variables by reference in Python, you can modify the properties of objects that were passed in. In my example (and others) you can modify members of the list that was passed in. You would not, however, be able to reassign the passed in variable entirely. For instance, see the following two pieces of code look like they might do something similar, but end up with different results:

def clear_a(x):
  x = []

def clear_b(x):
  while x: x.pop()

z = [1,2,3]
clear_a(z) # z will not be changed
clear_b(z) # z will be emptied

How do I pass a variable by reference?

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

the parameter passed in is actually a reference to an object (but the reference is passed by value)
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.

Python function calling on linked list, call by value/call by reference

It is a reference that is passed by value (i.e. it's similar to passing a pointer in C).

When you set root.next you are changing the value of next for the node that root refers to, and so the list changes at that node. When you set root itself, you are only modifying the reference that was passed in, not the underlying value that it refers to, and so the list is unaffected.

If you want to remove the last node of a linked list, you need to set the next of the second-to-last node to None. Something like:

def pop_last(root):
    if root is None or root.next is None:
        raise IndexError("can't pop from list with len < 2")
    if root.next.next is None:
        val = root.next.data
        root.next = None
        return val
    return pop_last(root.next)

How to pass function reference without using mutable type like list or dict

Functions are actually mutable, via __code__. Here's another example. However, don't know if this is a good idea.

do_something.__code__ = do_something_else.__code__
print(do_something())  # -> 1
print(foo())  # -> 1

Passing an integer by reference in Python

It doesn't quite work that way in Python. Python passes references to objects. Inside your function you have an object -- You're free to mutate that object (if possible). However, integers are immutable. One workaround is to pass the integer in a container which can be mutated:

def change(x):
    x[0] = 3

x = [1]
change(x)
print x

This is ugly/clumsy at best, but you're not going to do any better in Python. The reason is because in Python, assignment (=) takes whatever object is the result of the right hand side and binds it to whatever is on the left hand side *(or passes it to the appropriate function).

Understanding this, we can see why there is no way to change the value of an immutable object inside a function -- you can't change any of its attributes because it's immutable, and you can't just assign the "variable" a new value because then you're actually creating a new object (which is distinct from the old one) and giving it the name that the old object had in the local namespace.

Usually the workaround is to simply return the object that you want:

def multiply_by_2(x):
    return 2*x

x = 1
x = multiply_by_2(x)

*In the first example case above, 3 actually gets passed to x.__setitem__.

How can this be called Pass By Reference?

It is neither. It is call by sharing. I've also heard the term "pass by reference value" used.

Also known as "call by object" or "call by object-sharing," call by sharing is an evaluation strategy first named by Barbara Liskov et al. for the language CLU in 1974. It is used by languages such as Python, Iota, Java (for object references), Ruby, JavaScript, Scheme, OCaml, AppleScript, and many others. However, the term "call by sharing" is not in common use; the terminology is inconsistent across different sources. For example, in the Java community, they say that Java is call-by-value, whereas in the Ruby community, they say that Ruby is call-by-reference, even though the two languages exhibit the same semantics. Call by sharing implies that values in the language are based on objects rather than primitive types, i.e. that all values are "boxed".
The semantics of call by sharing differ from call by reference in that assignments to function arguments within the function aren't visible to the caller (unlike by reference semantics), so e.g. if a variable was passed, it is not possible to simulate an assignment on that variable in the caller's scope. However, since the function has access to the same object as the caller (no copy is made), mutations to those objects, if the objects are mutable, within the function are visible to the caller, which may appear to differ from call by value semantics. Mutations of a mutable object within the function are visible to the caller because the object is not copied or cloned — it is shared.

Python Functions Call by Reference