当存在可变引用时传递不可变引用

Question

我有一个 for 循环遍历一片 Point 结构。 Points 将在循环中修改一些字段，因此包含循环的函数需要对切片的可变引用。

当我需要将指向切片的（不可变）引用传递给迭代可变引用的 for 循环中的函数时，问题就出现了：

#[derive(Debug)]
struct Point {
    x: i32,
    y: i32,
}

fn main() {
    let mut grid = vec![];

    grid.push(Point { x: 10, y: 10 });
    grid.push(Point { x: -1, y: 7 });

    calculate_neighbors(&mut grid);
}

fn calculate_neighbors(grid: &mut [Point]) {
    for pt in grid.iter_mut() {
        pt.x = nonsense_calc(grid);
    }
}

#[allow(unused_variables)]
fn nonsense_calc(grid: &[Point]) -> i32 {
    unimplemented!();
}

Playground

error[E0502]: cannot borrow `*grid` as immutable because it is also borrowed as mutable
  --> src/main.rs:18:30
   |
17 |     for pt in grid.iter_mut() {
   |               ---------------
   |               |
   |               mutable borrow occurs here
   |               mutable borrow used here, in later iteration of loop
18 |         pt.x = nonsense_calc(grid);
   |                              ^^^^ immutable borrow occurs here

编译器抱怨 grid 不能作为不可变借用，因为可变借用已经存在。这是正确的，我可以看到它试图阻止的问题，但我该如何实现我需要做的事情呢？理想情况下，我不必创建 grid 的副本，因为这可能很昂贵。

Answer 1

避免为迭代借用数组的解决方案是使用索引：

fn calculate_neighbors(grid: &mut [Point]) {
    for i in 0..grid.len() {
        grid[i].x = nonsense_calc(grid);
    }
}

当存在可变引用时传递不可变引用

Passing an immutable reference when a mutable reference exists

loops

mutability

rust

borrow-checker