### 使用常量相關的方法 常量相關的方法允許我們為每個 enum 實例提供方法的不同實現，這使得常量相關的方法似乎是實現多路分發的完美解決方案。不過，通過這種方式，enum 實例雖然可以具有不同的行為，但它們仍然不是類型，不能將其作為方法簽名中的參數類型來使用。最好的辦法是將 enum 用在 switch 語句中，見下例： ```java // enums/RoShamBo3.java // Using constant-specific methods // {java enums.RoShamBo3} package enums; import static enums.Outcome.*; public enum RoShamBo3 implements Competitor<RoShamBo3> { PAPER { @Override public Outcome compete(RoShamBo3 it) { switch(it) { default: // To placate the compiler case PAPER: return DRAW; case SCISSORS: return LOSE; case ROCK: return WIN; } } }, SCISSORS { @Override public Outcome compete(RoShamBo3 it) { switch(it) { default: case PAPER: return WIN; case SCISSORS: return DRAW; case ROCK: return LOSE; } } }, ROCK { @Override public Outcome compete(RoShamBo3 it) { switch(it) { default: case PAPER: return LOSE; case SCISSORS: return WIN; case ROCK: return DRAW; } } }; @Override public abstract Outcome compete(RoShamBo3 it); public static void main(String[] args) { RoShamBo.play(RoShamBo3.class, 20); } } ``` 輸出為： ``` ROCK vs. ROCK: DRAW SCISSORS vs. ROCK: LOSE SCISSORS vs. ROCK: LOSE SCISSORS vs. ROCK: LOSE PAPER vs. SCISSORS: LOSE PAPER vs. PAPER: DRAW PAPER vs. SCISSORS: LOSE ROCK vs. SCISSORS: WIN SCISSORS vs. SCISSORS: DRAW ROCK vs. SCISSORS: WIN SCISSORS vs. PAPER: WIN SCISSORS vs. PAPER: WIN ROCK vs. PAPER: LOSE ROCK vs. SCISSORS: WIN SCISSORS vs. ROCK: LOSE PAPER vs. SCISSORS: LOSE SCISSORS vs. PAPER: WIN SCISSORS vs. PAPER: WIN SCISSORS vs. PAPER: WIN SCISSORS vs. PAPER: WIN ``` 雖然這種方式可以工作，但是卻不甚合理，如果采用 RoShamB02.java 的解決方案，那么在添加一個新的類型時，只需更少的代碼，而且也更直接。 :然而，RoShamBo3.java 還可以壓縮簡化一下： ```java // enums/RoShamBo4.java // {java enums.RoShamBo4} package enums; public enum RoShamBo4 implements Competitor<RoShamBo4> { ROCK { @Override public Outcome compete(RoShamBo4 opponent) { return compete(SCISSORS, opponent); } }, SCISSORS { @Override public Outcome compete(RoShamBo4 opponent) { return compete(PAPER, opponent); } }, PAPER { @Override public Outcome compete(RoShamBo4 opponent) { return compete(ROCK, opponent); } }; Outcome compete(RoShamBo4 loser, RoShamBo4 opponent) { return ((opponent == this) ? Outcome.DRAW : ((opponent == loser) ? Outcome.WIN : Outcome.LOSE)); } public static void main(String[] args) { RoShamBo.play(RoShamBo4.class, 20); } } ``` 輸出為： ``` PAPER vs. PAPER: DRAW SCISSORS vs. PAPER: WIN SCISSORS vs. PAPER: WIN SCISSORS vs. PAPER: WIN ROCK vs. SCISSORS: WIN ROCK vs. ROCK: DRAW ROCK vs. SCISSORS: WIN PAPER vs. SCISSORS: LOSE SCISSORS vs. SCISSORS: DRAW PAPER vs. SCISSORS: LOSE SCISSORS vs. ROCK: LOSE SCISSORS vs. ROCK: LOSE PAPER vs. ROCK: WIN PAPER vs. SCISSORS: LOSE SCISSORS vs. PAPER: WIN ROCK vs. SCISSORS: WIN SCISSORS vs. ROCK: LOSE SCISSORS vs. ROCK: LOSE SCISSORS vs. ROCK: LOSE SCISSORS vs. ROCK: LOSE ``` 其中，具有兩個參數的 compete() 方法執行第二個分發，該方法執行一系列的比較，其行為類似 switch 語句。這個版本的程序更簡短，不過卻比較難理解，對于一個大型系統而言，難以理解的代碼將導致整個系統不夠健壯。