`bool()` return type on with no arguments
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- Python
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Description
Right now bool has this __new__ signature: def __new__(cls: Type[_T], __o: object = ...) -> _T: ...
Should not it be:
@overload
def __new__(cls) -> Literal[False]: ...
@overload
def __new__(cls: Type[_T], __o: object) -> _T: ...
?
Current source:
https://github.com/python/typeshed/blob/2217ac8a8c7004baad94a5adcb0d503d985eaa21/stdlib/builtins.pyi#L676
What do you think?
Contributor guide
First steps
- Read the whole issue, then the project's contributing guide.
- Comment on the issue to say you are picking it up — it saves two people doing the same work.
- Fork the repository and make your change on a branch.
- Open a pull request that references the issue number.
Research direction
Start with stdlib/builtins.pyi around line 676 and read the issue's comment thread to resolve how bool() with and without an argument should be typed. Done means the agreed signature is represented in the stub and the resulting behavior is validated for both call forms.
Written by the indexing model from the issue text.
Assessment
- Tech stack
- python
- Domain
- tooling
- Issue type
- Bug
- Difficulty
- 3/5
- Estimated time
- 1-2 days
- Activity status
- Stale
- Clarity
- Clearly specified
- Newbie friendliness
- 45/100