How to detect that a function argument is the default one
Ben Finney
ben+python at benfinney.id.au
Thu Dec 11 18:03:28 EST 2014
Tony the Tiger <tony at tiger.invalid> writes:
> radius='10', mass='1'
>
> if radius == '10' ...
>
> if mass == '1' ...
This ignores the problem as stated: The OP wants to distinguish between
a value that was explicitly set by the caller, versus a value that was
set by default because the caller did not specify the parameter.
The general solution to this is to choose a value outside the domain, a
<URL:https://en.wikipedia.org/wiki/Sentinel_value> which is different
from any value the caller would choose for that parameter.
Often, the Python singleton ‘None’ is good for the purpose. I think in
this case, where the domain is numeric, a default of ‘None’ would do
fine.
class Foo:
def __init__(centre=(0, 0), radius=None, mass=None):
if radius is None:
# Caller didn't specify. Set a default value.
radius = 10
if mass is None:
# Caller didn't specify. Set a default value.
mass = 1
self.centre = centre
self.radius = radius
self.mass = mass
In some other cases, ‘None’ is a valid value for the caller to set, and
so cannot be used as a sentinel to distinguish the default. For those
cases, creating a specific object instance to serve as the sentinel is
simple and clear:
class Foo:
_SENTINEL = object()
def __init__(centre=(0, 0), radius=_SENTINEL, mass=_SENTINEL):
if radius is _SENTINEL:
# Caller didn't specify. Set a default value.
radius = 10
if mass is _SENTINEL:
# Caller didn't specify. Set a default value.
mass = 1
self.centre = centre
self.radius = radius
self.mass = mass
Creating a new object specifically for the sentinel means there is no
chance some other value will compare identical. Choosing a
leading-underscore name indicates to the reader that this is not part of
the public API.
--
\ “Odious ideas are not entitled to hide from criticism behind |
`\ the human shield of their believers' feelings.” —Richard M. |
_o__) Stallman |
Ben Finney
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