同一功能中的"活不够久"错误

Question

我希望这段代码能够正常工作,因为所有绑定都在同一范围内:

fn main() {
    let mut foobar = vec!["foo"];
    let bar = "bar".to_string();
    foobar.push(&bar);
}

但是我收到了这个错误:

error: `bar` does not live long enough
 --> baz.rs:4:18
  |>
4 |>     foobar.push(&bar);
  |>                  ^^^
note: reference must be valid for the block suffix following statement 0 at 2:33...
 --> baz.rs:2:34
  |>
2 |>     let mut foobar = vec!["foo"];
  |>                                  ^
note: ...but borrowed value is only valid for the block suffix following statement 1 at 3:32
 --> baz.rs:3:33
  |>
3 |>     let bar = "bar".to_string();
  |>                                 ^

error: aborting due to previous error

Answer 1

在同一块中声明的变量将按与声明它们相反的顺序删除.在你的代码中,bar删除之前foobar:

fn main() {
    let mut foobar = vec!["foo"]; // <---------| 0
    let bar = "bar".to_string();  // <--| 1    |
    foobar.push(&bar);            //    | bar  | foobar
                                  // <--|      |
                                  // <---------|
    // In the error message
    // 0 is called "block suffix following statement 0", and
    // 1 is called "block suffix following statement 1"
}

你是推引用到bar的foobar,所以你必须保证bar至少长达住的foobar.但是因为bar在之后宣布foobar,它bar的寿命实际上比foobars 短,这意味着foobar在短时间内包含一个悬空参考.

要使代码编译,请bar在之前声明foobar:

fn main() {
    let bar = "bar".to_string();
    let mut foobar = vec!["foo"];
    foobar.push(&bar);
}

或选择加入非词汇生命周期:

#![feature(nll)]

fn main() {
    let mut foobar = vec!["foo"];
    let bar = "bar".to_string();
    foobar.push(&bar);
}

虽然这仍然有一个悬空参考,但没关系,因为删除引用什么都不做; 在Vec不需要使用时,它的下降所包含的参考价值.