您如何摆脱对实现的接口的多余转换?

时间:2019-03-11 13:47:37

标签: java kotlin junit type-inference kotlintest

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的

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();
    }
}

我现在要对其进行测试(使用kotlintest,因为我对此很喜欢):

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
            } }
        }
    }
}

我需要强制转换为ICardSuit的地方,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as Array<out ICardSuit>,因为这是确保我仅访问界面属性的最简单方法,

但我真的不喜欢强制转换要测试的实例。

我能做些什么吗?

2 个答案:

答案 0 :(得分:3)

您是否有特定原因需要使用匹配器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,如果您确实想使用beSameInstanceAs,则可以:

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

我在这里并没有收到编译器的任何抱怨

答案 1 :(得分:2)

beTheSameInstanceAs(CardSuit.HEART)返回一个Matcher<CardSuit>,因此它不能匹配任意的ICardSuit。这是有道理的(尽管Matcher可能是 contra -variant,但在这里您需要协方差)。但是您可以:

  1. 显式调用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}")
    }
    

    并致电

    it should beTheSameInstanceAs(CardSuit.HEART).widen()
    

    (我认为类型推断应该在这里起作用)。

  3. 由于beTheSameInstanceAs(x)确实可以匹配任何内容,因此声明一个等效函数,该函数返回一个Matcher<Any>

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