CODEWITHSUNDEEP
rustownershipborrow-checker
Camden Narzt
Camden Narzt

Reputation: 2003

why doesn't this variable live long enough?

I'm trying to extract an optional arg from getopts, and getting a borrowed value doesn't live long enough error for the variable s.

code:

let cfgFilePath = match matches.opt_str("c") {
    Some(s) => Some(Path::new(&s.clone())),
    None => None
};

error:

main.rs:29:36: 29:45 error: borrowed value does not live long enough
main.rs:29         Some(s) => Some(Path::new(&s.clone())),
                                              ^~~~~~~~~
main.rs:31:7: 65:2 note: reference must be valid for the block suffix following statement 10 at 31:6...
main.rs:31     };
main.rs:32     let tmpdir = Path::new(&matches.opt_str("t").unwrap_or("/tmp/".to_string()));
main.rs:33     let name = matches.opt_str("n").unwrap_or_else(||{
main.rs:34         print_usage(&program, opts);
main.rs:35         panic!("error: -n NAME required");
main.rs:36     });
           ...

This happens regardless of .clone(), .to_owned(), .to_str() or anything else I've thought to try.

Upvotes: 3

Views: 175

Answers (1)

Steven
Steven

Reputation: 5780

Because Path::new(&x) returns an &Path that borrows it's contents from x.

Some(s) => Some(Path::new(&s.clone())), // Type is Option<&Path>
// reborrow --------------^

What you actually want to do is use a PathBuf (the owned equivalent of Path). PathBuf will take ownership of s instead of borrowing it.

let cfgFilePath = match matches.opt_str("c") {
    Some(s) => Some(PathBuf::from(s)),
    None => None
};

Upvotes: 5

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