在采用 &self 或 &mut self 的函数中进行模式匹配时，如何避免 ref 关键字?

问题描述：

Rust 书调用ref 关键字legacy".由于我想遵循隐含的建议来避免 ref，我该如何在下面的玩具示例中做到这一点?您也可以在 playground 上找到代码.

The Rust book calls the ref keyword "legacy". As I want to follow the implicit advice to avoid ref, how can I do it in the following toy example? You can find the code also on the playground.

struct OwnBox(i32);

impl OwnBox {
    fn ref_mut(&mut self) -> &mut i32 {
        match *self {
            OwnBox(ref mut i) => i,
        }

        // This doesn't work. -- Even not, if the signature of the signature of the function is
        // adapted to take an explcit lifetime 'a and use it here like `&'a mut i`.
        // match *self {
        //     OwnBox(mut i) => &mut i,
        // }

        // This doesn't work
        // match self {
        //     &mut OwnBox(mut i) => &mut i,
        // }
    }
}

答

由于 self 是 &mut Self 类型，所以匹配自身就足够了，而完全省略 ref.使用 *self 取消引用它或将 & 添加到匹配臂都会导致不必要的移动.

Since self is of type &mut Self, it is enough to match against itself, while omitting ref entirely. Either dereferencing it with *self or adding & to the match arm would cause an unwanted move.

fn ref_mut(&mut self) -> &mut i32 {
    match self {
        OwnBox(i) => i,
    }
}

对于像这样的新类型，&mut self.0 就足够了.

For newtypes such as this one however, &mut self.0 would have been enough.

这要归功于 RFC 2005 — 匹配人体工程学.

在采用 &self 或 &mut self 的函数中进行模式匹配时，如何避免 ref 关键字?

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