reference - 为什么我需要为返回引用的函数添加生命周期？

Question

我写了下面的代码:

const DIGIT_SPELLING: [&str; 10] = [
    "", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"
];

fn to_spelling_1(d: u8) -> &str {
    DIGIT_SPELLING[d as usize]
}

fn main() {
    let d = 1;
    let s = to_spelling_1(d);
    println!("{}", s);
}

这会产生以下编译器错误:

error[E0106]: missing lifetime specifier
 --> src/main.rs:5:28
  |
5 | fn to_spelling_1(d: u8) -> &str {
  |                            ^ expected lifetime parameter
  |
  = help: this function's return type contains a borrowed value with an elided lifetime, but the lifetime cannot be derived from the arguments
  = help: consider giving it an explicit bounded or 'static lifetime

为了解决这个问题，我将代码更改为:

const DIGIT_SPELLING: [&str; 10] = [
    "", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"
];

fn to_spelling_1<'a>(d: u8) -> &'a str { // !!!!! Added the lifetime. !!!!!
    DIGIT_SPELLING[d as usize]
}

fn main() {
    let d = 1;
    let s = to_spelling_1(d);
    println!("{}", s);
}

这段代码编译并运行没有错误。为什么我需要添加 'a 生命周期？为什么添加 'a 生命周期可以修复错误？

Answer 1

任何返回引用的函数都必须包含该引用的生命周期。如果该函数还采用至少一个引用参数，则 lifetime elision意味着您可以省略返回生命周期，但如果没有引用参数，则不会发生省略，就像您的示例中那样。

请注意，在您的情况下，使用显式 'static 生命周期而不是泛型会更有意义，因为您返回的值始终是 'static:

fn to_spelling_1(d: u8) -> &'static str {
    DIGIT_SPELLING[d as usize]
}