在发生改变的枚举中使用mem::{take(_), replace(_)}来保留所有值

Description

假定我们有一个&mut MyEnum，它有（至少）两个变体， A { name: String, x: u8 }和B { name: String }。 现在我们想如果x为零，把MyEnum::A改成B，同时保持MyEnum::B不变。

Example

use std::mem;

enum MyEnum {
    A { name: String, x: u8 },
    B { name: String }
}

fn a_to_b(e: &mut MyEnum) {
    if let MyEnum::A { name, x: 0 } = e {
        // this takes out our `name` and put in an empty String instead
        // (note that empty strings don't allocate).
        // Then, construct the new enum variant (which will
        // be assigned to `*e`).
        *e = MyEnum::B { name: mem::take(name) }
    }
}

这也适用于更多的变体：

use std::mem;

enum MultiVariateEnum {
    A { name: String },
    B { name: String },
    C,
    D
}

fn swizzle(e: &mut MultiVariateEnum) {
    use MultiVariateEnum::*;
    *e = match e {
        // Ownership rules do not allow taking `name` by value, but we cannot
        // take the value out of a mutable reference, unless we replace it:
        A { name } => B { name: mem::take(name) },
        B { name } => A { name: mem::take(name) },
        C => D,
        D => C
    }
}