改变从向量中获取的对象

Question

我无法解决一个明显的简单问题。

基本上我想将一个结构的实例推送到一个向量中，以便稍后将其取出并通过调用为该结构实现的函数来修改对象。

为了简化案例，我创建了以下测试代码，这反映了我对同样问题的看法。

    let mut strings = Vec::new();
    strings.push("Hello".to_string());
    let string_option = strings.last();
    let string = string_option.unwrap();
    string.shrink_to(1);

这里有编译错误

error[E0596]: cannot borrow `*string` as mutable, as it is behind a `&` reference
  --> src/main.rs:89:5
   |
88 |     let string = string_option.unwrap();
   |         ------ help: consider changing this to be a mutable reference: `&mut String`
89 |     string.shrink_to(1);
   |     ^^^^^^^^^^^^^^^^^^^ `string` is a `&` reference, so the data it refers to cannot be borrowed as mutable

然后我尝试了无穷无尽的变体，比如

    let mut strings = Vec::new();
    strings.push("Hello".to_string());
    let string_option = strings.last().as_mut();
    let string = string_option.unwrap();
    string.shrink_to(1);

...或...

    let mut strings = Vec::new();
    strings.push("Hello".to_string());
    let string_option = strings.last().as_deref_mut();
    let string = string_option.unwrap();
    string.shrink_to(1);

其实上面显示的代码是这段代码的简化，我本来想做的。

struct Bar {
    data: Vec<String>
}

impl Bar {
    fn shrink_first(&mut self) {

        let s_opt = self.data.last().as_mut();  // s_opt is of type Option<&mut &String>

        let s = s_opt.unwrap(); // s is of type &mut & String
        s.shrink_to(1);
    }
}

上面的代码带来了以下错误...

error[E0716]: temporary value dropped while borrowed
  --> src/main.rs:67:21
   |
67 |         let s_opt = self.data.last().as_mut();  // s_opt is of type Option<&mut &String>
   |                     ^^^^^^^^^^^^^^^^         - temporary value is freed at the end of this statement
   |                     |
   |                     creates a temporary which is freed while still in use
68 |
69 |         let s = s_opt.unwrap(); // s is of type &mut & String
   |                 ----- borrow later used here
   |
   = note: consider using a `let` binding to create a longer lived value

error[E0596]: cannot borrow `**s` as mutable, as it is behind a `&` reference
  --> src/main.rs:70:9
   |
70 |         s.shrink_to(1);
   |         ^^^^^^^^^^^^^^ cannot borrow as mutable

但在我看来，它总是有相同的根源，但我还没有想出该怎么办。

Answer 1

只需将 strings.last() 更改为 strings.last_mut()。

last() 方法 returns 标准（不可变）参考（更准确地说，Option<&T>）。

为了能够改变向量中的最后一个 String，您需要通过 last_mut().

获得一个 mutable 引用

改变从向量中获取的对象

Mutate an object which was taken from a vector

reference

mutable

rust

borrow