Without pointers, can I pass references as arguments in Python?
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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.
Python parameters are always "references".
The way parameters in Python works and the way they are explained on the docs can be confusing and misleading to newcomers to the languages, specially if you have a background on other languages which allows you to choose between "pass by value" and "pass by reference".
In Python terms, a "reference" is just a pointer with some more metadata to help the garbage collector do its job. And every variable and every parameter are always "references".
So, internally, Python pass a "pointer" to each parameter. You can easily see this in this example:
>>> def f(L):
... L.append(3)
...
>>> X = []
>>> f(X)
>>> X
[3]
The variable X points to a list, and the parameter L is a copy of the "pointer" of the list, and not a copy of the list itself.
Take care to note that this is not the same as "pass-by-reference" as C++ with the & qualifier, or pascal with the var qualifier.
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