存在可变引用时传递不可变引用

Question

我有一个for遍历一片Point结构的循环.的Point旨意在环路中修改了一些字段,因此含有该循环函数需要一个可变参考切片.

当我需要将指向切片的(不可变)引用传递给for循环中迭代可变引用的函数时,就会出现问题:

#[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!();
}

操场

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不能借用不可变的,因为已经存在可变的借位.这是正确的,我可以看到它试图阻止的问题,但我如何实现我需要做的事情？理想情况下,我不必创建副本grid,因为这可能很昂贵.

Answer 1

避免借用数组进行迭代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);
    }
}