Question

假设我有以下特征来定义接口并采用几个类型参数......

Foo

我想将伴侣对象用作特征的具体实现的工厂。我还想强制用户使用object Foo { def apply[A, B](thing: Thing): Foo[A, B] = { ??? } private case class FooImpl[A1, B1](thing: Thing) extends Foo[A1, B1] private case class AnotherFooImpl[A2, B1](thing: Thing) extends Foo[A2, B1] }接口而不是子类。所以我隐藏了伴随对象中的具体实现，如下所示：

val foo = Foo[A1, B1](thing)  // should be an instance of FooImpl

val anotherFoo = Foo[A2, B1](thing)  // should be an instance of AnotherFooImpl

我希望能够按如下方式使用工厂：

apply

如何实施public static void main(String[] args) { // Your scanner code here. int answer = processAndReturn(int1, int2); System.out.println("Larger Integer: " + answer); int lastDigit = processLargerInt(answer); System.out.print("Last Digit: " + lastDigit); } public static int processAndReturn(int int1, int int2){ return Math.max(int1, int2); } public static int processLargerInt(int answer) { return answer % 10; }方法来实现这一目标？这SO post似乎接近标记。

Answer 1

怎么样：

trait Foo[A, B]
trait Factory[A, B] {
  def make(thing: Thing): Foo[A, B]
}

class Thing

object Foo {
def apply[A, B](thing: Thing)(implicit ev: Factory[A, B]) = ev.make(thing)

private case class FooImpl[A, B](thing: Thing) extends Foo[A, B]
private case class AnotherFooImpl[A, B](thing: Thing) extends Foo[A, B]

implicit val fooImplFactory: Factory[Int, String] = new Factory[Int, String] {
  override def make(thing: Thing): Foo[Int, String] = new FooImpl[Int, String](thing)
}

implicit val anotherFooImplFactory: Factory[String, String] = new Factory[String, String] {
  override def make(thing: Thing): Foo[String, String] = new AnotherFooImpl[String, String](thing)
}

现在：

def main(args: Array[String]): Unit = {
  import Foo._

  val fooImpl = Foo[Int, String](new Thing)
  val anotherFooImpl = Foo[String, String](new Thing)

  println(fooImpl)
  println(anotherFooImpl)
}

收率：

FooImpl(testing.X$Thing@4678c730)
AnotherFooImpl(testing.X$Thing@c038203)

Answer 2

使用TypeTags（以克服类型参数的擦除），我们可以根据传递给apply方法的类型参数调用相应的隐藏实现，如下所示。它正确地实例化了各自的实现，但是Foo的类型信息丢失了，实际上它会像_202那样出现一些垃圾？我不知道为什么会发生这种情况以及如何为Foo保留正确的类型。也许有人可以对此有所了解。

trait Foo[A,B]
object Foo {
   def apply[A: TypeTag, B: TypeTag](thing: Thing) = 
    if(typeTag[A] == typeTag[Int])  
      FooImpl(thing) 
    else if(typeTag[A] == typeTag[String]) 
      AnotherFooImpl(thing) 
    else 
      new Foo[Double,Double] {}

   private case class FooImpl(thing: Thing) extends Foo[Int, String]
   private case class AnotherFooImpl(thing: Thing) extends Foo[String, String]
  } 

Foo[Int,String](new Thing) // Foo[_202, _203] = FooImpl($sess.cmd123$Thing@50350b75)

The actual types for _203 and _203 are: ???
// type _203 >: String with _201, type _202 >: Int with _200 


Foo[String,String](new Thing) //Foo[_202, _203] = AnotherFooImpl($sess.cmd123$Thing@51d80d6)

Scala工厂使用apply方法的泛型类型？

2 个答案: