memory-management - 使用 Option<&mut> 2 次

Question

下面的程序可以正常编译:

fn write_u16(bytes: &mut Vec<u8>, value: u16) {
    bytes.push((value >> 8) as u8);
    bytes.push(value as u8);
}

fn write_u32(bytes: &mut Vec<u8>, value: u32) {
    write_u16(bytes, (value >> 16) as u16);
    write_u16(bytes, value as u16);
}

现在我将字节类型更改为 Option<&mut Vec>:

fn write_u16(bytes_opt: Option<&mut Vec<u8>>, value: u16) {
    if let Some(bytes) = bytes_opt {
        bytes.push((value >> 8) as u8);
        bytes.push(value as u8);
    }
}

fn write_u32(bytes_opt: Option<&mut Vec<u8>>, value: u32) {
    write_u16(bytes_opt, (value >> 16) as u16);
    write_u16(bytes_opt, value as u16);
}

程序现在无法编译:

main.rs:10:15: 10:24 error: use of moved value: `bytes_opt`
main.rs:10     write_u16(bytes_opt, value as u16);
                         ^~~~~~~~~
main.rs:9:15: 9:24 note: `bytes_opt` moved here because it has type `core::option::Option<&mut collections::vec::Vec<u8>>`, which is non-copyable
main.rs:9     write_u16(bytes_opt, (value >> 16) as u16);
                        ^~~~~~~~~
error: aborting due to previous error

我不太明白，为什么我不能两次使用Option，我该如何解决这个问题？

我能想到的解决这个问题的唯一方法是:

fn write_u32(bytes_opt: Option<&mut Vec<u8>>, value: u32) {
    if let Some(bytes) = bytes_opt {
        write_u16(Some(bytes), (value >> 16) as u16);
        write_u16(Some(bytes), value as u16);
    } else {
        write_u16(None, (value >> 16) as u16);
        write_u16(None, value as u16);
    }
}

但这不是很好的代码。

Answer 1

虽然我怀疑在这种情况下你不应该这样做(正如已经评论过的那样)，但在这种情况下可以安全地重新借用可变引用:

fn write_u32(mut bytes_opt: Option<&mut Vec<u8>>, value: u32) {
    write_u16(bytes_opt.as_mut().map(|x| &mut **x), (value >> 16) as u16);
    write_u16(bytes_opt, value as u16);
}

bytes_opt.as_mut().map(|x| &mut **x) 也可以这样写 match bytes_opt { Some(&mut ref mut x) => Some(x) , 无 => 无, }。一个非常令人费解的模式(从左到右阅读它:&mut——解引用包含的值，ref mut——然后​​对其进行新的可变引用)，但是它有效并避免了所有权问题。