How to manually return a Result<(), Box<dyn Error>>?

Question

I want to return an error from a function in case a condition is true:

use std::error::Error;

pub fn run() -> Result<(), Box<dyn Error>> {
    // -- snip ---

    if condition {
        // return error
    }

    // -- snip --

    Ok(())
}

fn main() {}

I probably don't have the basics of the typesystem down, but everywhere I looked people use the ? operator, so I can't figure out what type to return.

1. Is it possible to just return an error like this?
2. Is there a better way to handle this logic?

Answer 1

Error is a trait and you want to return a trait object (note the dyn keyword), so you need to implement this trait:

use std::error::Error;
use std::fmt;

#[derive(Debug)]
struct MyError(String);

impl fmt::Display for MyError {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        write!(f, "There is an error: {}", self.0)
    }
}

impl Error for MyError {}

pub fn run() -> Result<(), Box<dyn Error>> {
    let condition = true;

    if condition {
        return Err(Box::new(MyError("Oops".into())));
    }

    Ok(())
}

fn main() {
    if let Err(e) = run() {
        println!("{}", e); // "There is an error: Oops"
    }
}

• Create your own error type,
• Implement Debug, Display, then Error for it,
• If there is an error, return the Err variant of Result.

I advise you to use failure that remove all the error boilerplate:

#[derive(Fail, Debug)]
#[fail(display = "There is an error: {}.", _0)]
struct MyError(String);

--

Note that if you expect an Error, you can return whatever type you want, given that it implements Error. This includes the error types in std.