rust - 在遍历对 Result 的引用时如何使用问号运算符？

Question

我需要分块我的矢量 Vec<Result<SomeStruct>>分成几 block ，遍历每一 block ，做一些工作，如果 Result<SomeStruct> 则返回错误包含它。

enum SomeStruct {}

#[derive(Debug)]
struct SomeError {}

type Result<T> = std::result::Result<T, SomeError>;

fn foo(v: Vec<Result<SomeStruct>>) -> Result<()> {
    for chunk in v.chunks(42) {
        for value in chunk {
            let value = value?;
            //do smth with value
        }
    }
    Ok(())
}

fn main() {
    let mut values = Vec::new();
    foo(values).unwrap();
}

playground

但是，我得到了错误

error[E0277]: the `?` operator can only be applied to values that implement `std::ops::Try`
  --> src/main.rs:11:25
   |
11 |             let value = value?;
   |                         ^^^^^^ the `?` operator cannot be applied to type `&std::result::Result<SomeStruct, SomeError>`
   |
   = help: the trait `std::ops::Try` is not implemented for `&std::result::Result<SomeStruct, SomeError>`
   = note: required by `std::ops::Try::into_result`

好像?期望值本身而不是对值的共享引用，但我无法取消引用共享引用并且不知道如何获取此值本身，因为 chunk这是 &[Result<SomeStruct>] , 和 value是&Result<SomeStruct>

我该如何解决？

Answer 1

一般情况

您的函数签名要求您返回一个拥有的 SomeError ，但您所拥有的只是对一个的引用。除非你有一些方法来转换 &SomeError进入SomeError ，这个问题是无解的。

最适用的解决方案是将您的错误更改为可克隆(或可复制，如果适用)然后转换您的 &Result<T, E>进入Result<&T, E>通过 Result::as_ref 和 Result::map_err :

#[derive(Debug, Clone)]
struct SomeError {}

for value in chunk {
    let value = value.as_ref().map_err(Clone::clone)?;
    // do something with `value`
}

也有可能，但不太常见的是，返回错误的引用。这要求错误的存在时间长于对函数的调用:

type Result<T, E = SomeError> = std::result::Result<T, E>;

fn foo(v: &[Result<SomeStruct>]) -> Result<(), &SomeError> {
    for chunk in v.chunks(42) {
        for value in chunk {
            let value = value.as_ref()?;
            // do something with `value`
        }
    }
    Ok(())
}

fn main() {
    let values = Vec::new();
    foo(&values).unwrap();
}

另见:

• How do I borrow a reference to what is inside an Option<T>?
• Why is it discouraged to accept a reference to a String (&String), Vec (&Vec), or Box (&Box) as a function argument?

具体案例

由于您拥有 Vec 的所有权并且不要利用内部迭代器是一个切片这一事实，您不需要 &Result开始。相反，转换 Vec进入迭代器和take大块长度的片段:

fn foo(v: Vec<Result<SomeStruct>>) -> Result<()> {
    let mut v = v.into_iter().peekable();

    while v.peek().is_some() {
        for value in v.by_ref().take(42) {
            let _value = value?;
            // do something with `value`
        }
    }

    Ok(())
}

您还可以使用 Itertools::chunks :

use itertools::Itertools; // 0.8.1

fn foo(v: Vec<Result<SomeStruct>>) -> Result<()> {
    for chunk in &v.into_iter().chunks(42) {
        for value in chunk {
            let value = value?;
            // do something with `value`
        }
    }
    Ok(())
}

另见:

• What is the canonical way to implement is_empty for Iterator?
• Are there equivalents to slice::chunks/windows for iterators to loop over pairs, triplets etc?
• Is it possible to split a vector into groups of 10 with iterators?