Please explain reasoning behind code snippet from argparse module

Question

I was looking at the source code for the built-in argparse._AppendAction, which implements the "append" action, and puzzled over the way it is implemented:

    def __call__(self, parser, namespace, values, option_string=None):
        items = _copy.copy(_ensure_value(namespace, self.dest, []))
        items.append(values)
        setattr(namespace, self.dest, items)

To break it down:

• _ensure_value is like dict.setdefault for attributes. That is, if namespace has an attribute with the name self.dest then it is returned, if not it is set to [] and returned.
• _copy.copy(x) returns just a shallow copy. When x is a list it is exactly like list(x) (but slower).
• Then the item is appended to the copy of the list gotten from namespace.
• Finally the self.dest attribute of namespace is overwritten with the copy, which should cause the old list to be garbage collected.

Why do it in such a roundabout and inefficient way, throwing away a whole list for each append? Why doesn't this suffice?

    def __call__(self, parser, namespace, values, option_string=None):
        items = _ensure_value(namespace, self.dest, [])
        items.append(values)

Answer 1

I'm not an expert in the implementation, so (disclaimer) this is really just a guess. With this implementation, the user can pass a list as a default=... in a call to add_argument without it being mutated within argparse. Perhaps this type of safety was desired by the developers for one reason or another.

The inefficiency you mention really isn't a big deal. It's for parsing commandline arguments, so this function is likely only called 10's of times per program under heavy usage.

---

I've tested this and indeed, If I use the following script (where argparse_temp is simply argparse.py copied to the current directory so I can play with it):

import argparse_temp as argparse

lst = [1,2,3]
parser = argparse.ArgumentParser()
parser.add_argument('-l',default=lst,action='append')
print parser.parse_args()
print lst

This prints (when called as: python test1.py -l 4):

Namespace(l=[1, 2, 3, '4'])
[1, 2, 3]

with argparse as is, but:

Namespace(l=[1, 2, 3, '4'])
[1, 2, 3, '4']

with your proposed change.

If you print the action that is returned by add_argument, you get:

_AppendAction(option_strings=['-l'], dest='l', nargs=None, const=None, default=[1, 2, 3, '4'], type=None, choices=None, help=None, metavar=None)

Which is it conceivable that argparse depends on that action elsewhere in the implementation. (Notice that default has been mutated here as well).