为什么我得到“必须使用的未使用结果......结果可能是一个 Err 变体，应该处理"?即使我正在处理它? [英] 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?

问题描述

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

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

结果:

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`

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推荐答案

您不是在处理结果，而是将结果从一种类型映射到另一种类型.

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

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

该行的作用是调用 foo()，它返回 Result<(), std::io::Error>.然后 map_err 使用您的闭包返回的类型(在本例中为 ())，并修改错误类型并返回 Result<(), ()>;.这是您没有处理的结果.由于您似乎只想忽略此结果，因此最简单的方法可能是调用 ok().

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() 将 Result 转换为 Option，将错误转换为 None>，您不会因为忽略它而收到警告.

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

或者:

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

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