metavar option has no effect when used with parser option if metavar set is the one generated without parser

[ArthurGarnier]

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Example Code

from typing import Annotated
import typer

app = typer.Typer()

def _foo(x: str):
    pass


@app.command()
def main(name: Annotated[str, typer.Argument(parser=_foo, metavar="NAME")]):
    pass


if __name__ == "__main__":
    app()

Description

When setting a parser to an Argument (or an Option) the metavar is replaced by the parser function name, in the above submitted snippet, if I don't set metavar, the metavar is automatically changed to _FOO in --help :

╰─$ python bug_typer.py --help
                                                                                                                                   
 Usage: bug_typer.py [OPTIONS] NAME                                                                                                
                                                                                                                                   
╭─ Arguments ─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────╮
│ *    name      _FOO  [default: None] [required]                                                                                 │
╰─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────╯

Maybe this is a root cause of the below issue.

So a workaroung to set a proper metavar is to set metavar="NAME", but by doing this, the metavar is not changed in the help output and stays at "_FOO" instead of "NAME". If I set anything else than "NAME", even "NAMe" it works as expected.

Operating System

Linux

Operating System Details

No response

Typer Version

0.16.0

Python Version

3.12.11

Additional Context

No response