Rust:将借用的结构传递给借用的枚举？

Question

我正在尝试将借用的结构传递到借用的枚举中。

#[derive(Copy, Clone)]
pub struct CustomerData {
    // Many fields about customers
}

#[derive(Copy, Clone)]
pub struct EmployeeData {
    // Many fields about employees
}

pub enum Person {
    Customer(CustomerData),
    Employee(EmployeeData)
}

fn do_something_with_customer(customer: &CustomerData) {
    let person = &Person::Customer(customer);

    // This would work, but this can be a large struct.
    // let person = &Person::Customer(customer.clone());

    general_method(person);
}

fn do_something_with_employee(employee: &EmployeeData) {
    let person = &Person::Employee(employee);

    // This would work, but this can be a large struct.
    // let person = &Person::Employee(employee.clone());

    general_method(person);
}

fn general_method(person: &Person) {

}

fn main() {
    let person = Person::Customer(CustomerData { });

    match &person {
        Person::Customer(data) => {
            do_something_with_customer(data);
        }
        Person::Employee(data) => {
            do_something_with_employee(data);
        }
    }
}

编译后得到的结果是：

error[E0308]: mismatched types
  --> src/main.rs:19:36
   |
19 |     let person = &Person::Customer(customer);
   |                                    ^^^^^^^^
   |                                    |
   |                                    expected struct `CustomerData`, found reference
   |                                    help: consider dereferencing the borrow: `*customer`
   |
   = note: expected type `CustomerData`
              found type `&CustomerData`

error[E0308]: mismatched types
  --> src/main.rs:28:36
   |
28 |     let person = &Person::Employee(employee);
   |                                    ^^^^^^^^
   |                                    |
   |                                    expected struct `EmployeeData`, found reference
   |                                    help: consider dereferencing the borrow: `*employee`
   |
   = note: expected type `EmployeeData`
              found type `&EmployeeData`

我知道Rust编译器不允许我这样做，但我觉得我应该可以这样做，因为我传递结构的枚举也是借来的。

此场景是否有模式/解决方案？也许使用Rc类型？在这种情况下，我可不想把代码弄得乱七八糟。

use std::rc::Rc;

#[derive(Copy, Clone)]
pub struct CustomerData {
    // Many fields about customers
}

#[derive(Copy, Clone)]
pub struct EmployeeData {
    // Many fields about employees
}

pub enum Person {
    Customer(Rc<CustomerData>),
    Employee(Rc<EmployeeData>)
}

fn do_something_with_customer(customer: Rc<CustomerData>) {
    let person = &Person::Customer(customer);

    // This would work, but this can be a large struct.
    // let person = &Person::Customer(customer.clone());

    general_method(person);
}

fn do_something_with_employee(employee: Rc<EmployeeData>) {
    let person = &Person::Employee(employee);

    // This would work, but this can be a large struct.
    // let person = &Person::Employee(employee.clone());

    general_method(person);
}

fn general_method(person: &Person) {

}

fn main() {
    let person = Person::Customer(Rc::new(CustomerData { }));

    match &person {
        Person::Customer(data) => {
            do_something_with_customer(data.clone());
        }
        Person::Employee(data) => {
            do_something_with_employee(data.clone());
        }
    }
}

Answer 1

您错误地识别了这个问题，编译器在其错误注释中说得很对。

您定义枚举的方式如下：

pub enum Person {
    Customer(CustomerData),
    Employee(EmployeeData)
}

但是您决定您的枚举成员应该是Person::Customer(&CustomerData)

fn do_something_with_customer(customer: &CustomerData) {
    let person = &Person::Customer(customer);

引用是不可传递的。因为&CustomerData是一个引用，并不意味着整个枚举将是对真实数据(即&Person::Customer(CustomerData))的引用。

有两种方法可以修复它；最明显的方法是查看CustomerData是否实现了Copy。如果是这样的话，你可以直接取消引用(并隐式复制)：

fn do_something_with_customer(customer: &CustomerData) {
    let person = Person::Customer(*customer);

(这是编译器的建议，所以我非常确定您的类型实现了Copy)

另一种选择是对类型执行#[derive(Clone)]并调用customer.clone()。再一次，以额外的分配为代价。

如果您确实需要枚举中的引用，则需要将枚举定义更改为：

pub enum Person<'a> {
    Customer(&'a CustomerData),
    Employee(&'a EmployeeData)
}

并处理对象属性现在是一个引用的事实，以及所有相关的问题。