如果条件满足,我如何查看矢量并弹出?

时间:2015-09-17 15:44:33

标签: rust

如果该元素的条件为真,我想从向量中检索元素。

fn draw() -> Option<String> {
    let mut v: Vec<String> = vec!["foo".to_string()];
    let t: Option<String>;
    let o = v.last();

    // t and v are actually a fields in a struct
    // so their lifetimes will continue outside of draw().

    match o {
        Some(ref e) => {
            if check(e) {
                t = v.pop();
                t.clone()
            } else {
                None
            }
        }
        None => None,
    }
}

fn check(e: &String) -> bool {
    true
}

fn main() {}

playground

这会导致错误:

error[E0502]: cannot borrow `v` as mutable because it is also borrowed as immutable
  --> src/main.rs:12:21
   |
4  |     let o = v.last();
   |             - immutable borrow occurs here
...
12 |                 t = v.pop();
   |                     ^ mutable borrow occurs here
...
20 | }
   | - immutable borrow ends here

我有点理解,但是我没有办法结束借阅(不使用clone())。

1 个答案:

答案 0 :(得分:8)

借用是词汇,因此v具有整个功能的生命周期。如果可以缩小范围,可以使用last()pop()。一种方法是使用功能样式的地图:

let mut v: Vec<String> = vec!["foo".to_string()];
if v.last().map_or(false, check) {
    v.pop()
} else {
    None
}