如何使以下代码类型安全？

Question

比方说,我有一个带有执行方法的通用命令trait,它接受一个I​​nput并返回一个Output.就像是

trait Input;
trait Output;

trait Command[I <: Input, O <: Output] {
  def execute(input: I): O;
}

然后,我将创建各种命令,类似于

class SampleInput extends Input
class SampleOutput extends Output

class SampleCommand extends Command[SampleInput, SampleOutput] {
  def execute(input:SampleInput):SampleOutput = new SampleOutput()
}

这里的问题是,我可以创建命令与SampleAInput和SampleBOutput编译器会接受愉快.我如何强制执行,以便编译器因类型不匹配错误而失败？

不知何故,我需要在类型下进行分组Input并Output传递该类型以创建命令.我怎么做？

Answer 1

trait InputOutput {
  type Input
  type Output
}

trait Command[IO <: InputOutput] {
  def execute(input: IO#Input): IO#Output
}

这是一些用法:

scala> trait SampleIO extends InputOutput {type Input = String; type Output = String}
defined trait SampleIO

scala> class SampleCommand extends Command[SampleIO] {def execute(input: String) = input}
defined class SampleCommand

scala> class SampleCommand extends Command[SampleIO] {def execute(input: String) = 1}
<console>:13: error: type mismatch;
 found   : Int(1)
 required: SampleIO#Output
       class SampleCommand extends Command[SampleIO] {def execute(input: String) = 1}
                                                                                 ^

Answer 2

由于您的约束是输入和输出的类型相同,我会尝试以下方法:

trait Input[T]
trait Output[T]

trait Command[T] {
  def execute[I <: Input[T], O <: Output[T]](i: I): O  
}

让我们尝试两种不同的类型.

class SampleInput extends Input[String]
class SampleOutput extends Output[Int]


scala> class SampleCommand extends Command[String] {                 
     |   def execute(input: SampleInput): SampleOutput = new SampleOutput
     | }
:10: error: class SampleCommand needs to be abstract, since method execute in trait Command of type [I <: Input[String],O <: Output[String]](i: I)O is not defined
       class SampleCommand extends Command[String] {
             ^