如何实现具有具体生命周期的 FromStr？

Question

我想为具有生命周期参数的结构实现 FromStr：

use std::str::FromStr;

struct Foo<'a> {
    bar: &'a str,
}

impl<'a> FromStr for Foo<'a> {
    type Err = ();
    fn from_str(s: &str) -> Result<Foo<'a>, ()> {

        Ok(Foo { bar: s })
    }
}

pub fn main() {
    let foo: Foo = "foobar".parse().unwrap();
}

然而，编译器抱怨：

error[E0495]: cannot infer an appropriate lifetime due to conflicting requirements
  --> src/main.rs:11:12
   |
11 |         Ok(Foo { bar: s })
   |            ^^^
   |
help: consider using an explicit lifetime parameter as shown: fn from_str(s: &'a str) -> Result<Foo<'a>, ()>
  --> src/main.rs:9:5
   |
9  |     fn from_str(s: &str) -> Result<Foo<'a>, ()> {
   |     ^

将实现更改为

impl<'a> FromStr for Foo<'a> {
    type Err = ();
    fn from_str(s: &'a str) -> Result<Foo<'a>, ()> {
        Ok(Foo { bar: s })
    }
}

给出这个错误

error[E0308]: method not compatible with trait
  --> src/main.rs:9:5
   |
9  |     fn from_str(s: &'a str) -> Result<Foo<'a>, ()> {
   |     ^ lifetime mismatch
   |
   = note: expected type `fn(&str) -> std::result::Result<Foo<'a>, ()>`
   = note:    found type `fn(&'a str) -> std::result::Result<Foo<'a>, ()>`
note: the anonymous lifetime #1 defined on the block at 9:51...
  --> src/main.rs:9:52
   |
9  |     fn from_str(s: &'a str) -> Result<Foo<'a>, ()> {
   |                                                    ^
note: ...does not necessarily outlive the lifetime 'a as defined on the block at 9:51
  --> src/main.rs:9:52
   |
9  |     fn from_str(s: &'a str) -> Result<Foo<'a>, ()> {
   |                                                    ^
help: consider using an explicit lifetime parameter as shown: fn from_str(s: &'a str) -> Result<Foo<'a>, ()>
  --> src/main.rs:9:5
   |
9  |     fn from_str(s: &'a str) -> Result<Foo<'a>, ()> {
   |     ^

Playpen

Answer 1

我认为您无法在这种情况下实施 FromStr。

fn from_str(s: &str) -> Result<Self, <Self as FromStr>::Err>;

特征定义中没有任何内容将输入的生命周期与输出的生命周期联系起来。

不是直接的答案，但我只是建议制作一个接受引用的构造函数：

struct Foo<'a> {
    bar: &'a str
}

impl<'a> Foo<'a> {
    fn new(s: &str) -> Foo {
        Foo { bar: s }
    }
}

pub fn main() {
    let foo = Foo::new("foobar"); 
}

这有没有任何故障模式的附带好处 - 不需要 unwrap。

你也可以只实现 From:

struct Foo<'a> {
    bar: &'a str,
}

impl<'a> From<&'a str> for Foo<'a> {
    fn from(s: &'a str) -> Foo<'a> {
        Foo { bar: s }
    }
}

pub fn main() {
    let foo: Foo = "foobar".into();
}

如何实现具有具体生命周期的 FromStr？

How do I implement FromStr with a concrete lifetime?

rust