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java - 你如何摆脱对实现接口的多余强制转换？

问题描述

假设我有一个界面

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

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

我决定用一个枚举来实现：

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，因为我很喜欢它）：

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
            } }
        }
    }
}

我需要转换到的地方，ICardSuit因为如果我不这样做，编译器会抱怨

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

我想保留它，as Array<out ICardSuit>因为这是确保我只访问接口属性的最简单方法，

但我真的不喜欢强制转换我正在测试的实例。

我能做些什么吗？

标签： javakotlinjunittype-inferencekotlintest

解决方案

您需要使用 matcher 是否有特定原因beSameInstanceAs？

您可以执行以下操作：

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
    }
}

我在这里并没有真正从编译器那里得到任何抱怨

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