[Python-ideas] Optional parameters without default value

[Serhiy Storchaka]

Function implemented in Python can have optional parameters with default 
value. It also can accept arbitrary number of positional and keyword 
arguments if use var-positional or var-keyword parameters (*args and 
**kwargs). But there is no way to declare an optional parameter that 
don't have default value. Currently you need to use the sentinel idiom 
for implementing this:

_sentinel = object()
def get(store, key, default=_sentinel):
     if store.exists(key):
         return store.retrieve(key)
     if default is _sentinel:
         raise LookupError
     else:
         return default

There are drawback of this:

* Module's namespace is polluted with sentinel's variables.

* You need to check for the sentinel before passing it to other function 
by accident.

* Possible name conflicts between sentinels for different functions of 
the same module.

* Since the sentinel is accessible outside of the function, it possible 
to pass it to the function.

* help() of the function shows reprs of default values. "foo(bar=<object 
object at 0xb713c698>)" looks ugly.


I propose to add a new syntax for optional parameters. If the argument 
corresponding to the optional parameter without default value is not 
specified, the parameter takes no value. As well as the "*" prefix means 
"arbitrary number of positional parameters", the prefix "?" can mean 
"single optional parameter".

Example:

def get(store, key, ?default):
     if store.exists(key):
         return store.retrieve(key)
     try:
         return default
     except NameError:
         raise LookupError

Alternative syntaxes:

* "=" not followed by an expression: "def get(store, key, default=)".

* The "del" keyword: "def get(store, key, del default)".

This feature is orthogonal to supporting positional-only parameters. 
Optional parameters without default value can be positional-or-keyword, 
keyword-only or positional-only (if the latter is implemented).