函数如何将值附加到向量并返回该值？

Question

我正在尝试编写一个接受矢量作为输入的函数rotate_card，将矢量的前元素旋转到后面，并返回一个包含旋转元素和旋转产生的矢量的对。

#[derive(Debug)]
enum Card {
    Ace,
    King,
    Queen,
    Jack,
}

type Deck = Vec<Card>;

fn rotate_card(deck: &mut Deck) -> (Card, &mut Deck) {
    let top_card = deck.remove(0);
    deck.push(top_card);
    (top_card, deck)
} // end rotate_card

fn main() {
    let mut my_deck: Deck = vec![Card::Ace, Card::King, Card::Queen, Card::Jack];
    let z: (Card, &mut Deck) = rotate_card(&mut my_deck);
    println!("The value of z is: {:?}.", z);
} // end main

error[E0382]: use of moved value: `top_card`
  --> src/main.rs:14:6
   |
13 |     deck.push(top_card);
   |               -------- value moved here
14 |     (top_card, deck)
   |      ^^^^^^^^ value used here after move
   |
   = note: move occurs because `top_card` has type `Card`, which does not implement the `Copy` trait

我如何解决value used after move错误？

Answer 1

我该如何解决？

你没有“解决”这类问题。所有权是Rust的基本概念，您必须了解它。

解决方案

Make your Card implicitly copyable via Copy

#[derive(Debug, Copy, Clone)]
enum Card { /* ... */ }

Make your Card explicitly copyable via Clone

#[derive(Debug, Clone)]
enum Card { /* ... */ }

fn rotate_card(deck: &mut Deck) -> Card {
    let top_card = deck.remove(0);
    deck.push(top_card.clone());
    top_card
}

Return a reference to the card

您可以将对最后一张卡而不是卡的引用作为值返回：

fn rotate_card(deck: &mut Deck) -> &mut Card {
    let top_card = deck.remove(0);
    deck.push(top_card);
    deck.last_mut().unwrap()
}

Use reference-counting

像Rc或Arc这样的类型可以允许共享值的所有权：

use std::rc::Rc;

type Deck = Vec<Rc<Card>>;

fn rotate_card(deck: &mut Deck) -> Rc<Card> {
    let top_card = deck.remove(0);
    deck.push(top_card.clone());
    top_card
}

其他说明

这是一个无用的函数签名：

fn rotate_card(deck: &mut Deck) -> (Card, &mut Deck) 

没有理由让Deck回到来电者那里;他们已经有了这个参考。删除它。

切片（和Vecs通过DerefMut）有rotate_left方法;你不需要重新实现它。