Simplify argument parsing default value

[VoidFunc]

Hi, I'm writing a new CLI tool and I'm using clap-rs for argument parsing.

In this tool you can specify in which port you wan't to listen TCP connections using the "-p PORT" flag. As you know, ports range is from 0 to 2^16. If the user put something out of this range I wan't to notify it through an error message and set port value to 8080.

This is my code:

My question is: Is there a better way to achieve this ?

[H2CO3]

You can slightly simplify this pattern using unwrap_or_else:

let port: u16 = p.parse().unwrap_or_else(|_| {
    eprintln!("specified port isn't in a valid range, setting to 8080");
    8080
});

---

Note 1: By the way, I would recommend against doing this. Specifying an invalid port number is an error, probably a typo or something, and falling back to a completely different (albeit valid) port number is going to be the source of many hours of head-banging against the table for the one running into this behavior. So I'd just report it as an error.

Note 2: Not sure how fine control you need over the arguments, but for cutting down on value extraction boilerplate, I'd suggest you to use structopt instead of directly using clap.

[VoidFunc]

Thank you very much for the tips and for the code simplification, I will keep them in mind.

Again, thank you!

[mbrubeck]

I agree with @H2CO3 that some broader changes might be best. But for the sake of comparison, here's a version that preserves all the original behavior but reduces some nesting and duplication. (Though I often find explicit match more readable than and/or combinators, so it's fine if you prefer the original code.)

let port: u16 = matches.value_of("port").and_then(|p|
    p.parse().ok().or_else(|| {
        eprintln!("specified port isn't in a valid range, setting to 8080");
        None
    })
).unwrap_or(8080);