Python: variables by references (hack)

Is there any way (hack) to push Python function (def) to return results by reference even for immutable types?

A proposal application (swap as subroutine):

def swap(a, b):

.....a,b = b,a

Note:

def swap(a, b):

.....return b,a

works as function which is not the answer of the question!

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For example there is a function random.shuffle(a) that works in-place. My idea is to c开发者_如何学Pythonall a function written in Fortran/C++ and call them via Python. It does work but has disadvantages too.

note:

Both "lambda" and "def" (as function) have the following problem: a, b = swap(a, b) which requires care about order of variables. In my proposal (if it was possible) the subroutine is used as: swap(a, b) so there is no requirement to care about order of variable.

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All names in Python are references. And no, there are no "out" references (e.g. in a C++ sense) available. You need to pass a mutable object and then you can mutate it in the function. But then again, returning new value(s) should be the preferred way.

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No, such things don't exist, because you get the given object as a reference, but if you re-assign it, it won't be changed back.

You either have to work with a mutable container (list, dict, object with attributes) in this case.

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In python there is no way to pass a "write pointer". You can pass an object and a name (so the callee can use for example setattr) or you can pass a list and an index. There is no such a thing as a context free "address of a modifiable cell"... there are only names or indexes that however also need a context (which namespace, which array).

In the general case if you really need to pass a write pointer a solution in Python is to pass a "setter" function that can be used by the callee to set the value.

For example:

def foo(setter):
    setter(42)

def bar():
    x = [1,2,3,4]

    def setFirst(value):
        x[0] = value

    foo(setFirst)

    print x[0] # --> 42