rust - 在 Rust 中返回提供的参数是惯用的吗？

Question

我有一些原型(prototype)代码:

impl MsgTrait for MsgA {
    fn apply_to(&self, state: State) -> State {
        match state {
            State::StateOne(mut state_one) => {
                state_one.common += 1; // just a mutability test
                State::StateOne(state_one)
            },
            _ => {
                state
            }
        }
    }
}

impl MsgTrait for MsgB {
    fn apply_to(&self, state: State) -> State {
        match state {
            State::StateOne(mut state_one) => {
                state_one.common += 2; // just a mutability test
                State::StateOne(state_one)
            },
            State::StateTwo(mut state_two) => {
                state_two.common += 3; // just a mutability test
                State::StateTwo(state_two)
            }
        }
    }
}

// this is a stub for receiving different kinds of messages from the network
fn recv() -> Msg {
    Msg::MsgA(Mega {field_a: 42})
}

fn main() {
    let mut state = State::StateOne(StateOne {common: 0, one_special: 1});
    for _ in 0..100 { // this would be loop, but that makes the playground timeout
        let incoming = recv(); // this would block
        match incoming {
            Msg::MsgA(msg_a) => {
                state = msg_a.apply_to(state)
            },
            Msg::MsgB(msg_b) => {
                state = msg_b.apply_to(state)
            }
        }
    }
}

https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=e7ddbbe51ce02c66dc3203bc2ecec104
为了变异 state并且仍然为循环的下一次迭代拥有它，我已经开始从方法中返回它。
这是 Rust 的惯用语吗？
如果我确实需要这样做，有没有办法避免重新包装 state_xxx在 State::StateXxx(state_xxx)在每种方法中？

Answer 1

如果您不想为每个突变重建状态，则可以考虑改为传递可变引用。例如
https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=74a8b1113204bc7abe31f9ced0851fd1