Missing default return value for typer.Option

[wu-clan]

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Example Code

from typing import Optional

import typer


def main(
        name: Optional[str] = typer.Option('Test', '--name', '-n'),
):
    if name:
        print(f"Good day Ms. {name}.")
    else:
        print(f"Hello World!")


if __name__ == "__main__":
    typer.run(main)

Description

When executing python scratch.py --name or python scratch.py -n, happend Option '-n' requires an argument., but it should detect the default value and return it instead of forcing a value to be filled in, if using typer.Argument , it doesn't have any problems

Operating System

Windows

Operating System Details

No response

Typer Version

0.6.1

Python Version

3.8.10

Additional Context

No response

[YuriiMotov]

The idea of default value is that it will be used if user doesn't specify this option at all

python scratch.py

python scratch.py -n will not work here, because Typer needs to know the exact number of values to expect for this option.
Otherwise we would have ambiguity: is the value after -n the value of this option or the value of another argument?