以下示例无法编译

let f = |x| {
    let _: &i32 = x;
    x
};
{
    let scoped = 42;
    f(&scoped);
} // <- scoped dropped here.
f(&42);

error[E0597]: `scoped` does not live long enough
  --> src/lib.rs:10:7
   |
10 |     f(&scoped);
   |       ^^^^^^^ borrowed value does not live long enough
11 | } // <- scoped dropped here.
   | - `scoped` dropped here while still borrowed
12 | f(&42);
   | - borrow later used here

For more information about this error, try `rustc --explain E0597`.

确实，该示例中f的签名是

impl Fn(&'inferred i32) -> &'inferred i32

其中'inferred代表某些尚未知的（需要推断的）但已确定的生命周期。

然后，

{
    let scoped = 42;
    f(&scoped); // `'inferred` must "fit" into this borrow…
} // <- and thus can't span beyond this point.
f(&42) // And yet `'inferred` is used here as well => Error!

解决方案是显式注释闭包签名中涉及的类型，更重要的是，该签名中涉及的**生命周期“空位”/占位符/参数**。

           // Rust sees this "hole" early enough in its compiler pass
           //                       to figure out that the closure signature
           // vv                    needs to be higher-order, **input-wise**
let f = |_x: &'_ i32| {
};
{
    let scoped = 42;
    f(&scoped);
}
f(&42);

这使得闭包签名的一侧实际上成为了高阶。不再是

impl Fn(&'inferred_and_thus_fixed i32)...

对于某个外部推断的（因此，固定的）生命周期'inferred_and_thus_fixed，我们现在有

impl for<'any> Fn(&'any i32)...

这可以工作，但**关于返回借用的问题是什么**？（同时保持高阶）

确实，以下示例无法编译

let f = |x: &'_ i32| -> &'_ i32 {
    x // <- Error, does not live long enough.
};

error: lifetime may not live long enough
 --> src/lib.rs:5:5
  |
4 | let f = |x: &'_ i32| -> &'_ i32 {
  |             -           - let's call the lifetime of this reference `'2`
  |             |
  |             let's call the lifetime of this reference `'1`
5 |     x // <- Error, does not live long enough.
  |     ^ returning this value requires that `'1` must outlive `'2`

原因是“显式生命周期‘空位’/占位符成为闭包签名中高阶生命周期参数”的机制仅适用于签名的一侧。

返回侧仍然使用推断的生命周期

let f = /* for<'any> */ |x: &'any i32| -> &'inferred i32 {
    x // <- Error, does not live long enough (when `'any < 'inferred`)
};

我们希望那里的f具有从函数签名的生命周期省略规则获得的签名，即

因此使用这个crate的原因。