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impl更改为

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

发出此错误

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

我不相信你可以在这种情况下实施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。

不是直接回答，但我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"); 
}

这样做的好处是没有任何失败模式 - 不需要unwrap。

您也可以实施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();
}

如何实现具有生命周期的FromStr？

1 个答案: