Value does not live long enough

问题

I don't completely understand lifetimes, but I think b's lifetime will end before self's.

So, how to edit this code? Do I copy something in memory? If I make a new instance, this lifetime must adhere to this case.

pub struct Formater {
    layout: &'static str,
}

impl Formater {
    pub fn new(layout: &'static str) -> Formater {
        let regex = Regex::new(r"%\{([a-z]+)(?::(.*?[^\\]))?\}").unwrap();
        let b = regex.replace_all(layout, "{}");

        return Formater {
            layout: &b,
        };
    }
}

The error:

error: `b` does not live long enough
  --> src/format.rs:16:22
   |
16 |             layout: &b,
   |                      ^ does not live long enough
17 |         };
18 |     }
   |     - borrowed value only lives until here
   |
   = note: borrowed value must be valid for the static lifetime...

回答1:

The scope of b is the new function, so its memory will be freed when the function returns. But you are trying to return a reference to b from that function. If Rust let you do this, the only code that could possibly use that reference would use it after the value is invalid. The borrow checker is protecting you from undefined behaviour.

Making layout to be &'static str sounds like you are making things simple, but it is unreasonable to expect the dynamically allocated memory from regex.replace_all to be static. Without getting into unsafe code, you should consider anything in the 'static lifetime to be a compile-time constant. For example, a string literal.

As others have said, you probably want layout to be a String. A String is similar to &str, but it owns the underlying str. That means when you move the String, the underlying str moves with it. A &str is a reference and must not outlive the owner of the str that it points to.

If you really want it to be &str, an alternative but less ergonomic approach is to have the caller of new() own the &str, and pass it in as a mutable reference.

pub struct Formatter<'a> {
    layout: &'a str,
}

impl <'a> Formatter<'a> {
    pub fn new(layout: &'a mut &str) -> Formatter<'a> {
        let regex = Regex::new(r"%\{([a-z]+)(?::(.*?[^\\]))?\}").unwrap();
        *layout = regex.replace_all(layout, "{}");

        return Formatter {
            layout: layout,
        };
    }
}

This moves the problem one layer up the call stack and means that the reference you pass to new will be mutated by new.

pub fn main() {
    let mut s = "blah %{blah}";
    {
        let formatter = Formatter::new(&mut s);
        println!("{:?}", formatter.layout); // "blah {}"
    }
    println!("{:?}", s); // "blah {}"
}

Now s is owned by main, so formatter is valid as long as it is used only in a smaller scope than main.

But overall, I think this method is messier and you should just stick with String unless you have a good reason.

来源：https://stackoverflow.com/questions/42503296/value-does-not-live-long-enough

标签

rust

lifetime

borrowing