rust - 借来的可变类型

Question

具有借来的可变引用的函数如何调用具有相同借来的可变引用的第二个函数？

fn main() {
    let mut a = vec![0, 1, 2, 3, 4];
    
    first_function(&mut a);
    
    println!("{:?}", a);
}

fn first_function(a: &mut Vec<i32>) {
    println!("...first function");
    a[0] = 5;
    second_function(&mut a);
}

fn second_function(a: &mut Vec<i32>) {
    println!("...second function");
    a[2] = 6;
}

编译器错误通常非常有帮助，但我不明白这一点。

error[E0596]: cannot borrow `a` as mutable, as it is not declared as mutable
  --> src/main.rs:12:21
   |
9  | fn first_function(a: &mut Vec<i32>) {
   |                   - help: consider changing this to be mutable: `mut a`
...
12 |     second_function(&mut a);
   |                     ^^^^^^ cannot borrow as mutable

...这是 Rust Playground中代码的链接

Answer 1

通过编写&mut a，您试图获取对a的可变引用，而不是对其值引用的可变引用，而是&mut &mut Vec<i32>。 a的值已经是您想要的可变引用:

second_function(a);

updated playground