Why am I getting "unused Result which must be used ... Result may be an Err variant, which should be handled" even though I am handling it?

Question

fn main() {
    foo().map_err(|err| println!("{:?}", err));
}

fn foo() -> Result<(), std::io::Error> {
    Ok(())
}

results:

warning: unused `std::result::Result` that must be used
 --> src/main.rs:2:5
  |
2 |     foo().map_err(|err| println!("{:?}", err));
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
  = note: #[warn(unused_must_use)] on by default
  = note: this `Result` may be an `Err` variant, which should be handled

    Finished dev [unoptimized + debuginfo] target(s) in 0.58s
     Running `target/debug/playground`

playground link

Answer 1

You're not handling the result, you're mapping the result from one type to another.

foo().map_err(|err| println!("{:?}", err));

What that line does is call foo(), which returns Result<(), std::io::Error>. Then map_err uses the type returned by your closure (in this case, ()), and modifies the error type and returns Result<(), ()>. This is the result that you are not handling. Since you seem to want to just ignore this result, the simplest thing to do would probably be to call ok().

foo().map_err(|err| println!("{:?}", err)).ok();

ok() converts Result<T,E> to Option<T>, converting errors to None, which you won't get a warning for ignoring.

Alternatively:

match foo() {
    Err(e) => println!("{:?}", e),
    _ => ()
}