Why is Vec missing lifetime specifier here? [duplicate]

问题

This code compiles:

struct IntDisplayable(Vec<u8>);

impl fmt::Display for IntDisplayable {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        for v in &self.0 {
            write!(f, "\n{}", v)?;
        }
        Ok(())
    }
}

fn main() {
        let vec: Vec<u8> = vec![1,2,3,4,5];
        let vec_Foo = IntDisplayable(vec);
        println!("{}",vec_Foo);
}

whilst this code doesn't:

struct StrDisplayable(Vec<&str>);

impl fmt::Display for StrDisplayable {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        for v in &self.0 {
            write!(f, "\n{}", v)?;
        }
        Ok(())
    }
}

fn main() {
        let vec: Vec<&str> = vec!["a","bc","def"];
        let vec_Foo = StrDisplayable(vec);
        println!("{}",vec_Foo);
}

error message:

error[E0106]: missing lifetime specifier
 --> src/lib.rs:3:27
  |
3 | struct StrDisplayable(Vec<&str>);
  |                           ^ expected lifetime parameter

What I'm trying to do is to implement fmt::Display for a Vec<&str>, which generally required wrapping Vec like this, however it only works for Vec<u8>, why substitute Vec<u8> into Vec<&str> led to such compile error?

回答1:

The compiler is told that you're borrowing a value, but not for how long it will live. Should it be static? Something else?

I presume you're trying to do the following.

struct StrDisplayable<'a>(Vec<&'a str>);

This way, you're explicitly telling the compiler that the strings will live at least as long as the struct, no less.

You'll also need to add in a lifetime in the implementation of the trait, which can by anonymous if using Rust 2018.

来源：https://stackoverflow.com/questions/57266481/why-is-vecstr-missing-lifetime-specifier-here

标签

rust

lifetime

Why is Vec<&str> missing lifetime specifier here? [duplicate]

问题

回答1: