以内容为条件的流行元素

Question

借用检查器让我很难找到一个优雅的实现。这是一个最小的例子：

use std::collections::VecDeque;

fn main() {
    let mut vec1 = VecDeque::new();
    vec1.push_back(2.);

    let mut vec2 = VecDeque::new();
    vec2.push_back(1.);

    while let (Some(x), Some(y)) = (vec1.front_mut(), vec2.front_mut()) {
        if x < y {
            *y -= *x;
            vec1.pop_front();
        } else {
            *x -= *y;
            vec2.pop_front();
        }
    }

    assert_eq!(vec2.len(), 0);
    assert_eq!(vec1.pop_front(), Some(1.));
}

这不会编译：

error[E0499]: cannot borrow `vec1` as mutable more than once at a time
  --> src/main.rs:13:13
   |
10 |     while let (Some(x), Some(y)) = (vec1.front_mut(), vec2.front_mut()) {
   |                                     ---- first mutable borrow occurs here
...
13 |             vec1.pop_front();
   |             ^^^^ second mutable borrow occurs here
...
18 |     }
   |     - first borrow ends here

error[E0499]: cannot borrow `vec2` as mutable more than once at a time
  --> src/main.rs:16:13
   |
10 |     while let (Some(x), Some(y)) = (vec1.front_mut(), vec2.front_mut()) {
   |                                                       ---- first mutable borrow occurs here
...
16 |             vec2.pop_front();
   |             ^^^^ second mutable borrow occurs here
17 |         }
18 |     }
   |     - first borrow ends here

if x < y 范围必须嵌套在 x 和 y 的范围中，那么我如何才能根据它们的第一个元素对 vec1/vec2 进行变异？

Answer 1

你可以这样做：

use std::collections::VecDeque;

fn main() {
    let mut vec1 = VecDeque::new();
    vec1.push_back(2.);

    let mut vec2 = VecDeque::new();
    vec2.push_back(1.);

    loop {
        let pop_vec1;

        if let (Some(x), Some(y)) = (vec1.front_mut(), vec2.front_mut()) {
            if x < y {
                *y -= *x;
                pop_vec1 = true;
            } else {
                *x -= *y;
                pop_vec1 = false;
            }
        } else {
            break;
        }

        if pop_vec1 {
            vec1.pop_front();
        } else {
            vec2.pop_front();
        }
    }

    assert_eq!(vec2.len(), 0);
    assert_eq!(vec1.pop_front(), Some(1.));
}

或者如果您更喜欢行数较少的循环：

loop {
    let pop_vec1 = match (vec1.front_mut(), vec2.front_mut()) {
        (Some(ref x), Some(ref mut y)) if x < y => { **y -= **x; true },
        (Some(x), Some(y)) => { *x -= *y; false },
        _ => break,
    };

    VecDeque::pop_front(if pop_vec1 { &mut vec1 } else { &mut vec2 });
}

不是很优雅，但很有效。

VecDeque 缺少方法 like Vec::drain_filter。也许 Rust 团队会添加它。