Question

我正在使用具有两种类型的demo 3库：

PhysicalDevice需要一个&Arc<Instance>。我正在编写一个函数，该函数将包含Arc的结构返回到这些类的实例，但是由于PhysicalDevice需要&Arc<Instance>的事实，借位检查程序未编译代码，即使实例超出了功能范围。

我对Pet取&Arc<Person>的问题进行了较小的复制。

在函数test中，我创建了两个对象，并尝试以结构体返回它们：

use std::sync::Arc;

struct Person {
    name: String,
}

struct Pet<'t> {
    name: String,
    owner: &'t Arc<Person>,
}

struct PersonPetPair<'a> {
    person: Arc<Person>,
    pet: Arc<Pet<'a>>,
}

fn test<'a>() -> PersonPetPair<'a> {
    let person = Person {
        name: String::from("ash"),
    };
    let person_arc = Arc::new(person);

    let pet = Pet {
        name: String::from("pikachu"),
        owner: &person_arc,
    };
    let pet_arc = Arc::new(pet);

    PersonPetPair {
        person: person_arc,
        pet: pet_arc,
    }
    // person_arc ends here, so the reference is invalid anymore
    // Still, the objects are still alive since they are created with
    // Arc<>
}

fn main() {
    let t = test();
    println!("{} owns {}", t.person.name, t.pet.name);
}

编译器告诉我有一个问题：

error[E0597]: `person_arc` does not live long enough
  --> src/main.rs:25:17
   |
25 |         owner: &person_arc,
   |                 ^^^^^^^^^^ borrowed value does not live long enough
...
36 | }
   | - borrowed value only lives until here
   |
note: borrowed value must be valid for the lifetime 'a as defined on the function body at 17:9...
  --> src/main.rs:17:9
   |
17 | fn test<'a>() -> PersonPetPair<'a> {
   |         ^^

实际上，函数结束时将删除局部变量person_arc。应该有一种避免此问题的方法，因为两个实例都由于使用了Arc而无法使用该功能。

引用超出功能范围的本地Arc

0 个答案: