Question

我有一个函数take_head，它有两个参数：一个切片，以及切片“head”中的项目数（“head”是第一个n项，而“尾巴”就是头后的一切。它将切片分为两部分：它返回的head和设置参数的tail。这是显示如何使用的main函数：

fn main() {
    let mut strings = &mut ["a", "b", "c"][..];
    println!("original: {:?}", strings);

    // head should ["a"], and strings should be set to the tail (["b", "c"]).
    let head = take_head(&mut strings, 1);

    println!("head: {:?}", head);    // Should print ["a"].
    println!("tail: {:?}", strings); // Should print ["b", "c"].
}

如果我这样实施take_head：

fn take_head<'a>(strings: &mut &'a mut [&'a str], n: usize) -> &'a mut [&'a str] {
    let value = std::mem::replace(strings, &mut []);
    let (head, tail) = value.split_at_mut(n);
    *strings = tail;

    println!("returning head: {:?}", head);
    head
}

it works correctly和输出：

original: ["a", "b", "c"]
returning head: ["a"]
head: ["a"]
tail: ["b", "c"]

但是，如果我像这样实施take_head：

// Make a convenient trait for slices.
pub trait TakeFrontMut<T> {
    fn take_front_mut(&mut self, n: usize) -> &mut [T];
}

impl<'a, T> TakeFrontMut <T> for &'a mut [T] {
    fn take_front_mut(&mut self, n: usize) -> &mut [T] {
        // It's the same code as before, just in a trait method.
        let value = std::mem::replace(self, &mut []);
        let (head, tail) = value.split_at_mut(n);
        *self = tail;
        return head;
    }
}

fn take_head<'a>(strings: &mut &'a mut [&'a str], n: usize) -> &'a mut [&'a str] {
    let head = strings.take_front_mut(n);
    println!("returning head: {:?}", head);
    head
}

It produces an error：

<anon>:15:24: 15:41 error: cannot infer an appropriate lifetime for autoref due to conflicting requirements [E0495]
<anon>:15     let head = strings.take_front_mut(n);
                                 ^~~~~~~~~~~~~~~~~
<anon>:14:1: 18:2 help: consider using an explicit lifetime parameter as shown: fn take_head<'a>(strings: &'a mut &'a mut [&'a str], n: usize)
 -> &'a mut [&'a str]
<anon>:14 fn take_head<'a>(strings: &mut &'a mut [&'a str], n: usize) -> &'a mut [&'a str] {
<anon>:15     let head = strings.take_front_mut(n);
<anon>:16     println!("returning head: {:?}", head);
<anon>:17     head
<anon>:18 }

问题：为什么第二个版本会产生错误？解析器无法确定合适的生命周期有什么不同？我不明白为什么会失败，我不确定这些相互矛盾的要求是什么。

_{是的，take_head函数是愚蠢的，但它是我能做的最简单的MVCE，它仍能捕获与我的真实代码相同的问题。}

Answer 1

take_front_mut的签名未指定返回值的正确生命周期。它应该是&'a mut [T]，因为这是您分割的切片的生命周期。这也要求你对特质本身进行改变。

pub trait TakeFrontMut<'a, T> {
    fn take_front_mut(&mut self, n: usize) -> &'a mut [T];
}

impl<'a, T> TakeFrontMut<'a, T> for &'a mut [T] {
    fn take_front_mut(&mut self, n: usize) -> &'a mut [T] {
        let value = std::mem::replace(self, &mut []);
        let (head, tail) = value.split_at_mut(n);
        *self = tail;
        return head;
    }
}

为什么Rust不能推断出我的特质方法的适当生命周期？

1 个答案: