如何将生命周期参数添加到不返回引用的闭包

Question

假设您有一个函数返回一个对引用起作用的闭包。当然，引用后面的对象必须至少在闭包被调用时存在。

这是一个非常简单的例子，它演示了问题。

fn get_cloned<'a>(obj: &'a MyStruct) -> impl Fn() -> MyStruct {
    || {obj.clone()}
}

编译器指出返回的闭包具有生命周期static：

cannot infer an appropriate lifetime
...but this borrow...rustc
main.rs(60, 56): this return type evaluates to the 'static lifetime...
main.rs(61, 5): ...but this borrow...

如何告诉编译器只要引用有效我就只使用函数的结果（闭包）？

谢谢！

[编辑]

您是否需要一个包含引用和闭包的虚拟结构？

struct Dummy<'a>{
  reference: &'a MyStruct,
  closure: Fn() -> MyStruct
}

?

假设克隆的成本非常高，并且可能永远不会调用闭包。 -> 惰性评估是必须的。

Answer 1

编译器告诉你该怎么做：

error: cannot infer an appropriate lifetime
 --> src/lib.rs:2:9
  |
1 | fn get_cloned<'a>(obj: &'a MyStruct) -> impl Fn() -> MyStruct {
  |                                         --------------------- this return type evaluates to the `'static` lifetime...
2 |         || {obj.clone()}
  |         ^^^^^^^^^^^^^^^^ ...but this borrow...
  |
note: ...can't outlive the lifetime 'a as defined on the function body at 1:15
 --> src/lib.rs:1:15
  |
1 | fn get_cloned<'a>(obj: &'a MyStruct) -> impl Fn() -> MyStruct {
  |               ^^
help: you can add a constraint to the return type to make it last less than `'static` and match the lifetime 'a as defined on the function body at 1:15
  |
1 | fn get_cloned<'a>(obj: &'a MyStruct) -> impl Fn() -> MyStruct + 'a {
  |                                         ^^^^^^^^^^^^^^^^^^^^^^^^^^

您需要在 return 类型上添加 + 'a。然后编译器会告诉你你缺少 move，但修复后，your code works:

fn get_cloned<'a>(obj: &'a MyStruct) -> impl Fn() -> MyStruct + 'a {
    || {obj.clone()}
}

如何将生命周期参数添加到不返回引用的闭包

How to add lifetime argument to closure not returning a reference

closures

lifetime

rust