New submission from Raymond Hettinger <raymond.hettin...@gmail.com>:
Python's help() function does not display overloaded function signatures.
For example, this code:
from typing import Union
class Smudge(str):
@overload
def __getitem__(self, index: int) -> str:
...
@overload
def __getitem__(self, index: slice) -> 'Smudge':
...
def __getitem__(self, index: Union[int, slice]) -> Union[str, 'Smudge']:
'Return a smudged character or characters.'
if isinstance(index, slice):
start, stop, step = index.indices(len(self))
values = [self[i] for i in range(start, stop, step)]
return Smudge(''.join(values))
c = super().__getitem__(index)
return chr(ord(c) ^ 1)
Currently gives this help:
__getitem__(self, index: Union[int, slice]) -> Union[str,
ForwardRef('Smudge')]
Return a smudged character or characters.
What is desired is:
__getitem__(self, index: int) -> str
__getitem__(self, index: slice) -> ForwardRef('Smudge')
Return a smudged character or characters.
The overload() decorator is sufficient for informing a static type checker but
insufficient for informing a user or editing tool.
----------
messages: 401052
nosy: rhettinger
priority: normal
severity: normal
status: open
title: Teach help about typing.overload()
_______________________________________
Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue45100>
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