为什么我需要向返回引用的函数添加生存期？

Question

我写了以下代码：

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);
}

这将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，我将我的代码更改为：

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时没有错误。为什么我需要添加'a生存期？为什么添加'a生存期可以修复该错误？

Answer 1

任何返回引用的函数都必须包含此引用的生存期。如果函数还接受至少一个按引用参数，则lifetime elision意味着您可以省略返回生命周期，但如果没有按引用参数，则不会发生省略，就像您的示例中一样。

请注意，在您的示例中，使用显式的'static生存期比使用泛型更有意义，因为您返回的值始终为'static

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