function - 局部函数访问外部变量

Question

A local function in Rust显示了如何制作不访问变量的局部函数。
该函数失败，因为它访问v:

// Assume v takes up 1 GB and is slow to compute
let v = [1,2,3];
fn recursive(x:usize) {
  if x - v.len() > 0 {
      // do stuff that may involve calling recursive() and change v
      recursive(x-1);
  }
}
recursive(8);

我该怎么做呢？

Answer 1

最简单的选择是通过(可变)引用将状态传递给递归函数:

let mut v = [1, 2, 3];
fn recursive(v: &mut [i32], x: usize) {
    if x - v.len() > 0 {
        // do stuff that may involve calling recursive() and change v
        recursive(v, x - 1);
    }
}
recursive(&mut v, 8);