你如何摆脱实现界面的多余演员？

Question

假设我有一个界面

public interface ICardSuit {
    /**short name*/
    public String getName();

    /** the colour of this card*/
    public ICardColour getColour();
}

我决定使用枚举实现：

public enum  CardSuit implements ICardSuit {
    HEART{
        @Override
        public ICardColour getColour() {
            return CardColour.RED;
        }
    },
    SPADE{
        @Override
        public ICardColour getColour() {
            return CardColour.BLACK;
        }
    },
    DIAMOND{
        @Override
        public ICardColour getColour() {
            return CardColour.RED;
        }
    },
    CLUBS {
        @Override
        public ICardColour getColour() {
            return CardColour.BLACK;
        }
    }
    ;

    @Override
    public String getName() {
        return this.name();
    }
}

我现在想测试它（使用kotlintest因为我正在开发它的喜好）：

class CardSuitTest : FunSpec(){
    init {
        test("there are exactly four suits"){CardSuit.values().size shouldBe 4}
        test("suits implement interface"){CardSuit.values().forEach { it shouldBe instanceOf(ICardSuit::class) }}
        test("suits have correct names"){
            val suits = CardSuit.values() as Array<out ICardSuit>
            suits.forEach { when(it.name){
                "HEART" -> it should beTheSameInstanceAs(CardSuit.HEART as ICardSuit)
                "SPADE" -> it should beTheSameInstanceAs(CardSuit.SPADE as ICardSuit)
                "DIAMOND" -> it should beTheSameInstanceAs(CardSuit.DIAMOND as ICardSuit)
                "CLUBS" -> it should beTheSameInstanceAs(CardSuit.CLUBS as ICardSuit)
            } }
        }
        test("suits have correct colours"){
            CardSuit.values().forEach { when(it){
                CardSuit.HEART,CardSuit.DIAMOND -> it.colour shouldBe CardColour.RED
                CardSuit.CLUBS, CardSuit.SPADE -> it.colour shouldBe CardColour.BLACK
            } }
        }
    }
}

我需要投射到ICardSuit，因为如果我不这样做，编译器就会抱怨

None of the following functions can be called with the arguments supplied.

* T.should(Matcher<T>)   where T cannot be inferred for    infix fun <T> T.should(matcher: Matcher<T>): Unit defined in io.kotlintest.matchers

* ICardSuit.should((ICardSuit) → Unit)   where T = ICardSuit for    infix fun <T> T.should(matcher: (T) → Unit): Unit defined in io.kotlintest.matchers

我想保留as Array<out ICardSuit>，因为这是确保我只访问界面属性的最简单方法，

但我真的不喜欢不得不施放我正在测试的实例。

我能做些什么吗？

Answer 1

是否有特定原因需要使用匹配器beSameInstanceAs？

你可以这样做：

val suits = CardSuit.values() as Array<out ICardSuite>

suits.forEach {
    when (it.name) {
        "HEART" -> it shouldBe CardSuit.HEART
        "SPADE" -> it shouldBe CardSuit.SPADE
   }
}

但是，如果您真的想使用beSameInstanceAs，您可以：

suits.forEach {
    when(it.name) {
       "HEART" -> it shouldBeSameInstanceAs CardSuit.HEART
       "SPADE" -> it shouldBeSameInstanceAs CardSuit.SPADE
    }
}

我真的没有收到编译器的任何抱怨

Answer 2

beTheSameInstanceAs(CardSuit.HEART)返回Matcher<CardSuit>，所以它无法匹配任意的ICardSuit。这是有道理的（虽然Matcher可能是反变体，你需要协方差）。但是你可以：

1. 明确地调用beTheSameInstanceAs<ICardSuit>(CardSuit.HEART)。
2. 创建一个辅助函数

   inline fun <T1, reified T2 : T1> Matcher<T2>.widen() = object : Matcher<T1>() {
       override fun test(value: T1) = 
           if (value is T2) 
               this.test(value) 
           else 
               Result(false, "$value is not a ${T2::class.name}", "$value is a ${T2::class.name}")
   }

   并打电话 it should beTheSameInstanceAs(CardSuit.HEART).widen() （我认为类型推断应该在这里工作）。
3. 因为beTheSameInstanceAs(x)真的可以匹配任何东西，声明一个返回Matcher<Any>的等效函数：

   fun beTheSameInstanceAsAny(x: Any) = beTheSameInstanceAs(x)
   
   // usage
   "HEART" -> it should beTheSameInstanceAsAny(CardSuit.HEART)