如何在Rust中给参考明确的寿命？

发布于 2025-02-10 04:33:18 字数 1266 浏览 1 评论 0 原文

我正在尝试返回 result＆lt;（），＆amp; str＆gt; 在Rust中，其中＆amp; str嵌入了有关发生的任何错误的数据。例如，说我有以下代码：

struct Foo {
    pub mynum :i32,
}

impl Foo {
    fn do_something(&self) -> Result<(), &str> {
        if self.mynum % 12 == 0 {
            return Err(&format!("error! mynum is {}", self.mynum))
        }
        Ok(())
    }
}

fn main() {
    let foo_instance = Foo{
        mynum: 36,
    };
    let result = foo_instance.do_something().unwrap();
    println!("{:?}",result)
}

如果我在 Rust Playground ，我得到的

error[E0515]: cannot return value referencing temporary value
 --> src/main.rs:9:20
  |
9 |             return Err(&format!("error! mynum is {}", self.mynum))
  |                    ^^^^^-----------------------------------------^
  |                    |    |
  |                    |    temporary value created here
  |                    returns a value referencing data owned by the current function

是不需要的。我如何告诉生锈编译器，用lifetime 'a 创建＆amp; str ？如果可能的话，我不想使用'static ，也不想加入 foo 带有额外的成员。

原文

I'm trying to return a Result<(), &str> in rust, where the &str has embedded data about any errors which have occurred. For example, say I have the following code:

struct Foo {
    pub mynum :i32,
}

impl Foo {
    fn do_something(&self) -> Result<(), &str> {
        if self.mynum % 12 == 0 {
            return Err(&format!("error! mynum is {}", self.mynum))
        }
        Ok(())
    }
}

fn main() {
    let foo_instance = Foo{
        mynum: 36,
    };
    let result = foo_instance.do_something().unwrap();
    println!("{:?}",result)
}

If I run this in the rust playground, I get

error[E0515]: cannot return value referencing temporary value
 --> src/main.rs:9:20
  |
9 |             return Err(&format!("error! mynum is {}", self.mynum))
  |                    ^^^^^-----------------------------------------^
  |                    |    |
  |                    |    temporary value created here
  |                    returns a value referencing data owned by the current function

which isn't desired.
How do I tell the Rust compiler that to create a &str with lifetime 'a where 'a is the lifetime of self? I don't want to use 'static if possible, and I don't want to load Foo with extra members..

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与君绝 2025-02-17 04:33:18

在这种情况下，您不应返回＆amp; str ，因为底层对象＆amp; str 在函数终止时将删除引用。本质上，您正在尝试返回对被删除的临时值的引用。请参阅 string之间的差异和＆amp; str 。

字符串可能是您想要的。此编译：

fn do_something(&self) -> Result<(), String> {
    if self.mynum % 12 == 0 {
        return Err(format!("error! mynum is {}", self.mynum));
    }
    Ok(())
}

You should not return a &str in this case because the underlying object the &str is referencing gets dropped when the function terminates. Essentially, you are attempting to return a reference to a temporary value which gets deleted. See this article on differences between String and &str.

String is probably what you want instead. This compiles:

fn do_something(&self) -> Result<(), String> {
    if self.mynum % 12 == 0 {
        return Err(format!("error! mynum is {}", self.mynum));
    }
    Ok(())
}

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