2

该练习要求我将特征实现到 Vec。测试在那里并且它们失败了,这是一个很好的起点。我已经完成了 String 的 trait 实现,这很容易,Vec 是另一回事。我不确定该方法需要返回什么,它在各种返回时都失败了。我正在提供原始代码、我的尝试以及我在尝试中遇到的错误。希望这就足够了。

Rustlings repo 的原始代码:

// traits2.rs
// 
// Your task is to implement the trait
// `AppendBar' for a vector of strings.
// 
// To implement this trait, consider for
// a moment what it means to 'append "Bar"'
// to a vector of strings.
// 
// No boiler plate code this time,
// you can do this!

// I AM NOT DONE

trait AppendBar {
    fn append_bar(self) -> Self;
}

//TODO: Add your code here




#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn is_vec_pop_eq_bar() {
        let mut foo = vec![String::from("Foo")].append_bar();
        assert_eq!(foo.pop().unwrap(), String::from("Bar"));
        assert_eq!(foo.pop().unwrap(), String::from("Foo"));
    }

}

以及我解决它的尝试:

// traits2.rs
//
// Your task is to implement the trait
// `AppendBar' for a vector of strings.
//
// To implement this trait, consider for
// a moment what it means to 'append "Bar"'
// to a vector of strings.
//
// No boiler plate code this time,
// you can do this!

// I AM NOT DONE
use std::clone::Clone;
trait AppendBar {
    fn append_bar(&mut self) -> Self;
}

//TODO: Add your code here
impl<T: Clone> AppendBar for Vec<T> {
    fn append_bar(&mut self) -> Self {
        let bar: T = String::from("Bar");
        self.to_vec().push(bar)
        // self.to_vec()
    }
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn is_vec_pop_eq_bar() {
        let mut foo = vec![String::from("Foo")].append_bar();
        assert_eq!(foo, vec![String::from("Foo"), String::from("Bar")]);
        assert_eq!(foo.pop().unwrap(), String::from("Bar"));
        assert_eq!(foo.pop().unwrap(), String::from("Foo"));
    }
}

编译为错误:


! Compiling of exercises/traits/traits2.rs failed! Please try again. Here's the output:
error[E0308]: mismatched types
  --> exercises/traits/traits2.rs:22:22
   |
20 | impl<T: Clone> AppendBar for Vec<T> {
   |      - this type parameter
21 |     fn append_bar(&mut self) -> Self {
22 |         let bar: T = String::from("Bar");
   |                  -   ^^^^^^^^^^^^^^^^^^^ expected type parameter `T`, found struct `std::string::String`
   |                  |
   |                  expected due to this
   |
   = note: expected type parameter `T`
                      found struct `std::string::String`
   = help: type parameters must be constrained to match other types
   = note: for more information, visit https://doc.rust-lang.org/book/ch10-02-traits.html#traits-as-parameters

error[E0308]: mismatched types
  --> exercises/traits/traits2.rs:23:9
   |
21 |     fn append_bar(&mut self) -> Self {
   |                                 ---- expected `std::vec::Vec<T>` because of return type
22 |         let bar: T = String::from("Bar");
23 |         self.to_vec().push(bar)
   |         ^^^^^^^^^^^^^^^^^^^^^^^ expected struct `std::vec::Vec`, found `()`
   |
   = note: expected struct `std::vec::Vec<T>`
           found unit type `()`

error: aborting due to 2 previous errors

For more information about this error, try `rustc --explain E0308`.

我已经阅读并重新阅读了书中建议的部分和特征,但这超出了我的范围。我确定这是一个简单的解决方案,但我看不到它。

4

4 回答 4

3

阿洛索的回答有出入。安德烈也给了。

当你接受self

fn append_bar(self) -> Self {
    self.push("Bar".to_owned());
    self
}

你正在接受一个可变的Vec

let mut foo = vec![String::from("Foo")].append_bar();
assert_eq!(foo.pop().unwrap(), String::from("Bar"));
assert_eq!(foo.pop().unwrap(), String::from("Foo"));

即使变量foo被声明为可变的,该方法也append_bar()接受一个不可变的变量。您不需要借用self,因为您没有尝试获得全部所有权,而是尝试修改驻留在所述变量中的现有数据。正确答案是

fn append_bar(mut self) -> Self {
    self.push("Bar".to_owned()); // || .to_string() || String::from("Bar") 
    // Whatever gets the point across. As the String literal is essentially a "Borrowed" string.
    self
}

append_bar()您尝试改变Strings 的集合并将其与附加的字符串一起返回的范围内。

于 2021-01-04T16:35:31.463 回答
2

有几个问题:

  • 您尝试将 a 推String送到 generic Vec<T>,其中T可以是任何类型!
  • 方法签名与赋值不同:您的方法定义为
    fn append_bar(&mut self) -> Self
    
    但应该是
    fn append_bar(self) -> Self
    
  • 您尝试返回 的结果Vec::push,但此方法不返回任何内容。

要解决第一个问题,请实现特征 forVec<String>而不是Vec<T>. 这就是任务所要求的:

// Your task is to implement the trait
// `AppendBar' for a vector of strings.

要解决第二个问题,您必须删除&,因此该方法接受拥有的值。

要解决最后一个问题,请在单独的语句中self调用后返回:Vec::push

self.push(bar);
self
于 2020-03-20T14:55:48.463 回答
1

我相信这个问题的正确答案应该是这样的:

trait AppendBar {
    fn append_bar(self) -> Self;
}

impl AppendBar for Vec<String> {
    fn append_bar(mut self) -> Self {
        self.push("Bar".to_string());
        self
    }
}

于 2020-12-05T14:45:42.587 回答
0

感谢@Aloso 的帮助和 Jussi,我也设法使示例正常工作。

为了编译需要突变,所以我最终得到了编译如下的代码:


// traits2.rs
//
// Your task is to implement the trait
// `AppendBar' for a vector of strings.
//
// To implement this trait, consider for
// a moment what it means to 'append "Bar"'
// to a vector of strings.
//
// No boiler plate code this time,
// you can do this!

// I AM NOT DONE

trait AppendBar {
    fn append_bar(&mut self) -> Self;
}

//TODO: Add your code here
impl AppendBar for Vec<String> {
    fn append_bar(&mut self) -> Self {
        self.push(String::from("Bar"));
        self.to_vec()
    }
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn is_vec_pop_eq_bar() {
        let mut foo = vec![String::from("Foo")].append_bar();
        assert_eq!(foo, vec![String::from("Foo"), String::from("Bar")]);
        assert_eq!(foo.pop().unwrap(), String::from("Bar"));
        assert_eq!(foo.pop().unwrap(), String::from("Foo"));
    }
}

于 2020-03-20T15:47:04.870 回答