“借入的价值不能活得足够长"具有返回impl trait的泛型函数 [英] "borrowed value does not live long enough" with a generic function that returns impl trait

问题描述

我从此Rust代码收到意外错误:

I get an unexpected error from this Rust code:

struct Container<'a> {
    x: &'a i32,
}

trait Reply {}
impl Reply for i32 {}

fn json<T>(_val: &T) -> impl Reply {
    3
}

fn f() -> impl Reply {
    let i = 123;
    let a = Container { x: &i };
    json(&a)
}

游乐场

错误是:

error[E0597]: `i` does not live long enough
  --> src/lib.rs:14:28
   |
12 | fn f() -> impl Reply {
   |           ---------- opaque type requires that `i` is borrowed for `'static`
13 |     let i = 123;
14 |     let a = Container { x: &i };
   |                            ^^ borrowed value does not live long enough
15 |     json(&a)
16 | }
   | - `i` dropped here while still borrowed

为什么?

如果将 json()的声明更改为以下两个版本之一，则代码将编译:

If I change the declaration of json() to either of these versions, the code compiles:

fn json(val: &Container) -> impl Reply

fn json<T>(val: &T) -> i32

只有在同时具有类型参数和返回的特征对象的情况下，编译器才会拒绝代码.

It is only when there is both a type parameter and a returned trait object that the compiler rejects the code.

这是我们对 warp :: reply :: json()遇到的一个实际问题的减少，但是我更希望从总体上理解它.

This is a reduction from a real issue we had with warp::reply::json(), but I would prefer to understand it in general.

推荐答案

当参数和函数的返回类型是通用的时，Rust编译器假定返回类型可能会借用参数.这就是为什么它假定 f()返回一个引用局部变量 i 的值.

When the arguments and the return type of a function are generic, the Rust compiler assumes that the return type could, potentially, borrow the arguments. That's why it assumes that f() returns a value referencing the local variable i.

我不确定，但是我认为这是理想的，因为有人可以针对可能会出现问题的类型实现 Reply .

I'm not entirely sure, but I think this is desired, because someone could implement Reply for a type where this would be problematic.

由于存在错误，因此无法使用.已经在GitHub上进行了报道.

This doesn't work because of a bug. It has already been reported on GitHub.

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