function - Without pointers, can I pass references as arguments in Python?

Since Python doesn't have pointers, I am wondering how I can pass a reference to an object through to a function instead of copying the entire object. This is a very contrived example, but say I am writing a function like this:

def some_function(x):
    c = x/2 + 47
    return c

y = 4
z = 12

print some_function(y)
print some_function(z)

From my understanding, when I call some_function(y), Python allocates new space to store the argument value, then erases this data once the function has returned c and it's no longer needed. Since I am not actually altering the argument within some_function, how can I simply reference y from within the function instead of copying y when I pass it through? In this case it doesn't matter much, but if y was very large (say a giant matrix), copying it could eat up some significant time and space.

Answer

Your understanding is, unfortunately, completely wrong. Python does not copy the value, nor does it allocate space for a new one. It passes a value which is itself a reference to the object. If you modify that object (rather than rebinding its name), then the original will be modified.

Edit

I wish you would stop worrying about memory allocation: Python is not C++, almost all of the time you don't need to think about memory.

It's easier to demonstrate rebinding via the use of something like a list:

def my_func(foo):
    foo.append(3)  # now the source list also has the number 3
    foo = [3]      # we've re-bound 'foo' to something else, severing the relationship
    foo.append(4)  # the source list is unaffected
    return foo


original = [1, 2]
new = my_func(original)


print original     # [1, 2, 3]
print new          # [3, 4]

It might help if you think in terms of names rather than variables: inside the function, the name "foo" starts off being a reference to the original list, but then we change that name to point to a new, different list.