Question

Possible Duplicate:
Understanding Python's call-by-object style of passing function arguments

I recently came across this:

x = [1,2,3]

def change_1(x):
    x = x.remove(x[0])
    return x

Which results in:

>>> change_1(x)
>>> x
[2, 3]

I find this behavior surprising, as I thought that whatever goes in inside a function has no effect on outside variables. Furthermore, I constructed an example where basically do the same thing but without using remove:

x = [1,2,3]

def change_2(x):
    x = x[1:]
    return x

Which results in:

>>> change_2(x)
[2, 3] # Also the output prints out here not sure why this is
>>> x
[1, 2, 3]

And I get the result that I would expect, the function does not change x.

So it must be something remove specific that has effect. What is going on here?

Answer 1

One of the things that is confusing is that you’ve called lots of different things ‘x’. For example:

def change_1(x):       # the x parameter is a reference to the global 'x' list
    x = x.remove(x[0]) # on the left, though, is a new 'x' that is local to the function
    return x           # the local x is returned

>>> x = [1, 2, 3]
>>> y = change_1(x)    # assign the return value to 'y'
>>> print y
None                   # this is None because x.remove() assigned None to the local 'x' inside the function
>>> print x
[2, 3]                 # but x.remove() modified the global x inside the function

def change_2(x):
    x = x[1:]          # again, x on left is local, it gets a copy of the slice, but the 'x' parameter is not changed
    return x           # return the slice (copy)

>>> x = [1, 2, 3] 
>>> y = change_2(x)
>>> print x
[1, 2, 3]             # the global 'x' is not changed!
>>> print y
[2, 3]                # but the slice created in the function is assigned to 'y'