为什么不能将引用的生命周期推断为所有上下文可能性中最短的？

Question

这是一个代码示例，我在其中测试了两个 &str 和 return 其中之一：

fn bad_longest(s1: &str, s2: &str) -> &str {
  if s1.len() >= s2.len() { s1 } else { s2 }
}

它没有编译请求显式生命周期，所以我提供了它们：

fn longest<'r, 'a, 'b>(s1: &'a str, s2: &'b str) -> &'r str
where
  'a: 'r,
  'b: 'r
{
  if s1.len() >= s2.len() { s1 } else { s2 }
}

现在，以下测试顺利通过：

static STATIC: &str = "123";

fn main() {
  let auto = "123456";
  let dyn_ = String::from("123456789");
  println!(
    "{}",
    longest(
      longest(STATIC, auto),
      dyn_.as_str()
    )
  );
}

我的问题是：我手动提供的生命周期不是可以从上下文中明显推导出来的吗？我是否遗漏了任何用例？

Answer 1

函数签名中省略的生命周期永远不会从它们的使用方式中推断出来。还有一些simple rules for inferring elided lifetimes，完全基于签名本身：

Each elided lifetime in the parameters becomes a distinct lifetime parameter.

If there is exactly one lifetime used in the parameters (elided or not), that lifetime is assigned to all elided output lifetimes.

In method signatures there is another rule

If the receiver has type &Self or &mut Self, then the lifetime of that reference to Self is assigned to all elided output lifetime parameters.

您的函数有两个非 self 参数，因此 none 这些规则可以为 return 值提供生命周期，因此需要明确的生命周期。

为什么不能将引用的生命周期推断为所有上下文可能性中最短的？

Why can't the lifetime of a reference be deduced as the shortest of all the contextual possibilities?

lifetime

object-lifetime

rust

template-argument-deduction

return-type-deduction