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What is unwrap in Rust, and what is it used for?

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rust

I have this code that uses .unwrap():

fn main() {     let paths = std::fs::read_dir("/home/user").unwrap();      for path in paths {         println!("Name: {}", path.unwrap().path().display());      } } 

After looking at the definition of unwrap,

pub fn unwrap(self) -> T {   match self {         Ok(t) => t,         Err(e) => unwrap_failed("called `Result::unwrap()` on an `Err` value", e),     } } 

And the signature of read_dir

pub fn read_dir<P: AsRef<Path>>(path: P) -> io::Result<ReadDir> 

Am I correct in understanding that unwrap returns the T type that is passed in Result?

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Angel Angel Avatar asked Apr 01 '16 17:04

Angel Angel


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1 Answers

In Rust, when you have an operation that may either return a T or fail, you will have a value of type Result<T,E> or Option<T> (E will be the error condition in case of an interesting error).

The function unwrap(self) -> T will give you the embedded T if there is one. If instead there is not a T but an E or None then it will panic.

It is best used when you are positively sure that you don't have an error. If that is not the case usually it is better either pattern-match the error or use the try! macro ? operator to forward the error.

In your example, the call to read_dir() returns a io::Result<ReadDir> because opening the directory might fail. And iterating the opened directory returns multiple values of type io::Result<DirEntry> because reading the directory might also fail.

With try! ? it would be something like this:

fn try_main() -> std::io::Result<()> {     let entries = std::fs::read_dir("/home/user")?;      for entry in entries {         println!("Name: {}", entry?.path().display());      }     Ok(()) }  fn main() {     let res = try_main();      if let Err(e) = res {         println!("Error: {}", e);     } } 

Look how every error case is checked.

(Updated to use ? instead of try!(). The macro still works, but the ? is preferred for new code).

like image 118
rodrigo Avatar answered Oct 13 '22 07:10

rodrigo