rust - 如何链接采用先前结果并满足严格顺序的函数？

Question

示例:

let response = add_customer(InputCustomer)
    .validate()?
    .generate_code()
    .create(DB::create(pool_conextion))?;

我尝试使用各种结构，但我不知道这是否是最好的方法:

struct InputCustomer {}

fn add_customer(i: InputCustomer) -> Validate {
    Validate {
        result: InputCustomer {},
    }
}

struct Validate {
    result: InputCustomer,
}

impl Validate {
    fn do_validate() -> GenCode {
        // valdiate struct customer
        GenCode {
            result: InputCustomer {},
        }
    }
}

struct GenCode {
    result: InputCustomer,
}

impl GenCode {
    fn generate_code() -> Create {
        // generate customer code
        Create { result: true }
    }
}

struct Create {
    result: bool,
}

Answer 1

您可以使用phantom type parameters在单个结构上实现所有功能。 。 Customer 结构保存一些状态:

pub struct Customer<State> {
    state: PhantomData<State>,
}

我们可以创建客户可能处于的状态:

pub struct CustomerStateNew;
pub struct CustomerStateValidated;
pub struct CustomerStateWithCode;

当您创建客户时，它的状态为CustomerStateNew:

pub fn add_customer() -> Customer<CustomerStateNew> {
    Customer { state: PhantomData }
}

要验证客户，它必须处于CustomerStateNew状态:

impl Customer<CustomerStateNew> {
    pub fn validate(&self) -> Customer<CustomerStateValidated> {
        Customer { state: PhantomData }
    }
}

必须验证客户 (CustomerStateValidated) 才能生成代码:

impl Customer<CustomerStateValidated> {
    pub fn generate_code(&self) -> Customer<CustomerStateWithCode> {
        Customer { state: PhantomData }
    }
}

并且它必须具有要创建的生成代码 (CustomerStateWithCode)。 create 消耗 self，因此客户在创建后就无法使用(您可能不希望出现这种行为，但为了完整起见，我将其包含在此处):

impl Customer<CustomerStateWithCode> {
    pub fn create(self) -> Result<(), ()> {
        Ok(())
    }
}

现在我们可以将创建用户的方法链接在一起:

let result = add_customer().validate().generate_code().create()?;

但是，如果我们在验证之前尝试创建Customer，则代码将无法编译:

let result = add_customer().create();

// error[E0599]: no method named `create` found for struct `Customer<CustomerStateNew>`
//   --> src/main.rs:36:20
// 36 |     add_customer().create();
//   |                    ^^^^^^ method not found in `Customer<CustomerStateNew>`

此外，其他人都无法创建具有任意状态的 Customer，因为 state 字段是私有(private)的:

mod somewhere_else {
    fn bla() {
        let customer: Customer<CustomerStateWithCode> = Customer { state: PhantomData };
        customer.create();
    }
}

// error[E0451]: field `state` of struct `Customer` is private
//    --> src/main.rs:41:64
//    |
// 41 |  let customer: Customer<CustomerStateWithCode> = Customer { state: PhantomData };
//    |    

如果您想要存储特定于每个状态的数据，您可以将实际的State 存储在Customer 中，而不是PhantomData。然而现在，状态不仅仅是编译时安全，而且将在运行时存储:

pub struct CustomerStateWithCode(pub usize);

pub struct Customer<State> {
    state: State,
}

impl Customer<CustomerStateValidated> {
    pub fn generate_code(&self) -> Customer<CustomerStateWithCode> {
        Customer { state: CustomerStateWithCode(1234) }
    }
}

我们使用幻像类型创建了一个简单的状态机。这也称为类型状态模式。请注意，状态将被编译为空，因此没有运行时成本，只有编译时安全!

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