Theo Julienne <theo.julie...@gmail.com> added the comment:
def list_again(foo):
foo.append("bar")
def list_again_again(foo):
foo = foo + ["1"]
The part that can be confusing is that 'foo' is a *copy* of a *reference* to an
object. Because 'foo' is a copy, assigning to it will only alter a local copy,
while calling methods on it will always call on the original object (because
it's a reference to the same place). When an object is mutable, it just has no
methods that modify it, so the only way to change it is by assignment, which
will never result in changes to the outside function. There are no special
cases, but it does require understanding pass-by-(copy/reference) and object
mutability, and is against most people's intuitions.
The way that behaves is actually how most *programmers* would expect (because
they are used to it), though. Every other language I can think of does it this
way. For example, in C, a pointer is still a copy of a reference -- if you
assign to the pointer, it wont propagate up to the caller. The same goes in
Java.
Perhaps what makes it harder to understand in Python is that everything is an
object and looks the same regardless of mutability, whereas other languages
typically have conventions (such as capitalising: int, str, Dict, List) to make
it clearer when something is an object (which usually means the same as a
python mutable object) as opposed to a built-in type (which usually means the
same as a python immutable object).
Just my 2c
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nosy: +Theo.Julienne
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Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue9702>
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