为什么Vec& str>这里缺少生命周期说明符?

问题描述

此代码编译:

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);
}

此代码没有:

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);
}

错误消息:

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

我想做的是为 Vec<& str> 实施 fmt :: Display ，这通常需要包装 Vec ，例如此，但是它仅适用于 Vec< u8> ，为什么将 Vec< u8> 替换为 Vec<& str> 会导致这种编译错误?

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?

推荐答案

编译器被告知您正在借用一个值，但是没有使用该值多久.它应该是静态的吗?还有吗?

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

我想您正在尝试执行以下操作.

I presume you're trying to do the following.

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

这样，您就明确地告诉编译器，这些字符串的生存时间至少与该结构一样长.

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

您还需要延长特质的实现期限，如果使用Rust 2018，则可以匿名添加.

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

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