为什么我会收到错误“未实现特征‘迭代器’"?对泛型类型的引用,即使它实现了 IntoIterator? [英] Why do I get the error "the trait `Iterator` is not implemented" for a reference to a generic type even though it implements IntoIterator?

查看:34
本文介绍了为什么我会收到错误“未实现特征‘迭代器’"?对泛型类型的引用,即使它实现了 IntoIterator?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在下面的示例中,MyTrait 扩展了 IntoIterator,但在循环中使用时编译器无法识别它.

In the following example, MyTrait extends IntoIterator but the compiler doesn't recognize it when used in a loop.

pub trait MyTrait: IntoIterator<Item = i32> {
    fn foo(&self);
}

pub fn run<M: MyTrait>(my: &M) {
    for a in my {
        println!("{}", a);
    }
}

我收到错误:

error[E0277]: `&M` is not an iterator
 --> src/lib.rs:6:14
  |
6 |     for a in my {
  |              ^^ `&M` is not an iterator
  |
  = help: the trait `Iterator` is not implemented for `&M`
  = note: required because of the requirements on the impl of `IntoIterator` for `&M`
  = note: required by `into_iter`

推荐答案

只有 M 实现了 IntoIterator,但您正在尝试迭代 &M,这不是必须的.

Only M implements IntoIterator, but you're trying to iterate over a &M, which doesn't have to.

不清楚您希望通过 run 实现什么,但删除引用可能是一个开始:

It's not clear what you hope to achieve with run, but removing the reference might be a start:

pub fn run<M: MyTrait>(my: M) {
    for a in my {
        println!("{}", a);
    }
}

请注意,M 本身 可能是(或包含)一个引用,因此以这种方式编写它并不意味着您不能将其与借用数据一起使用.这是使用 run 迭代 &Vec (playground):

Note that M itself may be (or contain) a reference, so writing it in this way doesn't mean you can't use it with borrowed data. Here's one way to use run to iterate over a &Vec (playground):

impl<I> MyTrait for I
where
    I: IntoIterator<Item = i32>,
{
    fn foo(&self) {}
}

fn main() {
    let v = vec![10, 12, 20];
    run(v.iter().copied());
}

这使用 .copied()Iterator 转换为 Iterator.

This uses .copied() to turn an Iterator<Item = &i32> to an Iterator<Item = i32>.

这篇关于为什么我会收到错误“未实现特征‘迭代器’"?对泛型类型的引用,即使它实现了 IntoIterator?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆