Typer subcommand when called as function call from another subcommand doesn't take default typer option value, but it takes the typer object as parameter.

[pmaru10]

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Example Code

import typer

app = typer.Typer()


@app.command()
def cmd1(number: int = typer.Option(1, help="An integer for cmd1")):
    typer.echo(f"Running cmd1 with number: {number}")
    cmd2()

@app.command()
def cmd2(number: int = typer.Option(2, help="An integer for cmd2")):
    typer.echo(f"Running cmd2 with number: {number}")


if __name__ == "__main__":
    app()
~

Description

Typer subcommand when called as a function call from another subcommand, the function doesn't take the default value as an argument but rather takes the typer object.

In the above sample snippet, i was expecting the output

$ python3 test.py cmd1
Running cmd1 with number: 1
Running cmd2 with number: 2
$ 

But, I am seeing the following output.

$ python3 test.py cmd1
Running cmd1 with number: 1
Running cmd2 with number: <typer.models.OptionInfo object at 0x7fa1a5c29a90>
$ 

Am I missing here something?

Operating System

Linux

Operating System Details

Ubuntu 20.04

Typer Version

0.7.0

Python Version

Python 3.8.10

Additional Context

No response

[Jylpah]

Hi,

I do not know how Typer works internally, but you called cmd2() so the default argument will be typer.Option(2, help="An integer for cmd2") as you have specified in the function definition. If you call cmd2(2) it should do what you expected. I guess that Typer wrapper deal with the function parameter setting. So I do not see any bug here.