我想采取一个现有的enum,并添加更多的元素,如下所示:

enum A {a,b,c}

enum B extends A {d}

/*B is {a,b,c,d}*/

这在Java中可行吗?


当前回答

我的编码方式如下:

// enum A { a, b, c }
static final Set<Short> enumA = new LinkedHashSet<>(Arrays.asList(new Short[]{'a','b','c'}));

// enum B extends A { d }
static final Set<Short> enumB = new LinkedHashSet<>(enumA);
static {
    enumB.add((short) 'd');
    // If you have to add more elements:
    // enumB.addAll(Arrays.asList(new Short[]{ 'e', 'f', 'g', '♯', '♭' }));
}

LinkedHashSet既提供了每个条目只存在一次,又提供了它们的顺序保留。如果顺序无关紧要,则可以使用HashSet。以下代码在Java中是不可行的:

for (A a : B.values()) { // enum B extends A { d }
    switch (a) {
        case a:
        case b:
        case c:
            System.out.println("Value is: " + a.toString());
        break;
        default:
            throw new IllegalStateException("This should never happen.");
    }
}

代码可以这样写:

for (Short a : enumB) {
    switch (a) {
        case 'a':
        case 'b':
        case 'c':
            System.out.println("Value is: " + new String(Character.toChars(a)));
        break;
        default:
            throw new IllegalStateException("This should never happen.");
    }
}

从Java 7开始,你甚至可以用String做同样的事情:

// enum A { BACKWARDS, FOREWARDS, STANDING }
static final Set<String> enumA = new LinkedHashSet<>(Arrays.asList(new String[] {
        "BACKWARDS", "FOREWARDS", "STANDING" }));

// enum B extends A { JUMP }
static final Set<String> enumB = new LinkedHashSet<>(enumA);
static {
    enumB.add("JUMP");
}

使用enum替换:

for (String a : enumB) {
    switch (a) {
        case "BACKWARDS":
        case "FOREWARDS":
        case "STANDING":
            System.out.println("Value is: " + a);
        break;
        default:
            throw new IllegalStateException("This should never happen.");
    }
}

其他回答

enum A {a,b,c}
enum B extends A {d}
/*B is {a,b,c,d}*/

可以写成:

public enum All {
    a       (ClassGroup.A,ClassGroup.B),
    b       (ClassGroup.A,ClassGroup.B),
    c       (ClassGroup.A,ClassGroup.B),
    d       (ClassGroup.B) 
...

ClassGroup.B.getMembers()包含{a,b,c,d}

如何使用:假设我们想要这样的东西: 我们有事件,我们使用枚举。可以通过类似的处理对这些枚举进行分组。如果我们有多个元素的操作,那么有些事件开始操作,有些只是步骤,有些则结束操作。为了收集这样的操作和避免长开关情况,我们可以将它们分组,如示例所示,并使用:

if(myEvent.is(State_StatusGroup.START)) makeNewOperationObject()..
if(myEnum.is(State_StatusGroup.STEP)) makeSomeSeriousChanges()..
if(myEnum.is(State_StatusGroup.FINISH)) closeTransactionOrSomething()..

例子:

public enum AtmOperationStatus {
STARTED_BY_SERVER       (State_StatusGroup.START),
SUCCESS             (State_StatusGroup.FINISH),
FAIL_TOKEN_TIMEOUT      (State_StatusGroup.FAIL, 
                    State_StatusGroup.FINISH),
FAIL_NOT_COMPLETE       (State_StatusGroup.FAIL,
                    State_StatusGroup.STEP),
FAIL_UNKNOWN            (State_StatusGroup.FAIL,
                    State_StatusGroup.FINISH),
(...)

private AtmOperationStatus(StatusGroupInterface ... pList){
    for (StatusGroupInterface group : pList){
        group.addMember(this);
    }
}
public boolean is(StatusGroupInterface with){
    for (AtmOperationStatus eT : with.getMembers()){
        if( eT .equals(this))   return true;
    }
    return false;
}
// Each group must implement this interface
private interface StatusGroupInterface{
    EnumSet<AtmOperationStatus> getMembers();
    void addMember(AtmOperationStatus pE);
}
// DEFINING GROUPS
public enum State_StatusGroup implements StatusGroupInterface{
    START, STEP, FAIL, FINISH;

    private List<AtmOperationStatus> members = new LinkedList<AtmOperationStatus>();

    @Override
    public EnumSet<AtmOperationStatus> getMembers() {
        return EnumSet.copyOf(members);
    }

    @Override
    public void addMember(AtmOperationStatus pE) {
        members.add(pE);
    }
    static { // forcing initiation of dependent enum
        try {
            Class.forName(AtmOperationStatus.class.getName()); 
        } catch (ClassNotFoundException ex) { 
            throw new RuntimeException("Class AtmEventType not found", ex); 
        }
    }
}
}
//Some use of upper code:
if (p.getStatus().is(AtmOperationStatus.State_StatusGroup.FINISH)) {
    //do something
}else if (p.getStatus().is(AtmOperationStatus.State_StatusGroup.START)) {
    //do something      
}  

添加一些更高级的内容:

public enum AtmEventType {

USER_DEPOSIT        (Status_EventsGroup.WITH_STATUS,
              Authorization_EventsGroup.USER_AUTHORIZED,
              ChangedMoneyAccountState_EventsGroup.CHANGED,
              OperationType_EventsGroup.DEPOSIT,
              ApplyTo_EventsGroup.CHANNEL),
SERVICE_DEPOSIT     (Status_EventsGroup.WITH_STATUS,
              Authorization_EventsGroup.TERMINAL_AUTHORIZATION,
              ChangedMoneyAccountState_EventsGroup.CHANGED,
              OperationType_EventsGroup.DEPOSIT,
              ApplyTo_EventsGroup.CHANNEL),
DEVICE_MALFUNCTION  (Status_EventsGroup.WITHOUT_STATUS,
              Authorization_EventsGroup.TERMINAL_AUTHORIZATION,
              ChangedMoneyAccountState_EventsGroup.DID_NOT_CHANGED,
              ApplyTo_EventsGroup.DEVICE),
CONFIGURATION_4_C_CHANGED(Status_EventsGroup.WITHOUT_STATUS,
              ApplyTo_EventsGroup.TERMINAL,
              ChangedMoneyAccountState_EventsGroup.DID_NOT_CHANGED),
(...)

在上面,如果我们有一些失败(myEvent.is(State_StatusGroup.FAIL)),然后通过之前的事件迭代,我们可以很容易地检查我们是否必须通过以下方法恢复资金转移:

if(myEvent2.is(ChangedMoneyAccountState_EventsGroup.CHANGED)) rollBack()..

它可以用于:

包括关于处理逻辑的显式元数据,不需要记住 实现一些多继承 我们不想使用类结构,例如发送简短的状态消息

我自己也有同样的问题,我想把我的观点发表出来。我认为这样做有几个激励因素:

您希望有一些相关的枚举代码,但在不同的类中。在我的例子中,我有一个基类,在一个相关的枚举中定义了几个代码。在以后的某一天(今天!)我想为基类提供一些新功能,这也意味着枚举的新代码。 派生类既支持基类的枚举,也支持它自己的枚举。没有重复的enum值!如何为子类创建一个枚举,包括父类的枚举及其新值。

使用接口并不能真正解决问题:您可能会意外地获得重复的enum值。不可取的。

我最终只是组合了枚举:这确保了不会有任何重复的值,但代价是与相关类的绑定不那么紧密。但是,我认为重复的问题是我主要担心的……

我的编码方式如下:

// enum A { a, b, c }
static final Set<Short> enumA = new LinkedHashSet<>(Arrays.asList(new Short[]{'a','b','c'}));

// enum B extends A { d }
static final Set<Short> enumB = new LinkedHashSet<>(enumA);
static {
    enumB.add((short) 'd');
    // If you have to add more elements:
    // enumB.addAll(Arrays.asList(new Short[]{ 'e', 'f', 'g', '♯', '♭' }));
}

LinkedHashSet既提供了每个条目只存在一次,又提供了它们的顺序保留。如果顺序无关紧要,则可以使用HashSet。以下代码在Java中是不可行的:

for (A a : B.values()) { // enum B extends A { d }
    switch (a) {
        case a:
        case b:
        case c:
            System.out.println("Value is: " + a.toString());
        break;
        default:
            throw new IllegalStateException("This should never happen.");
    }
}

代码可以这样写:

for (Short a : enumB) {
    switch (a) {
        case 'a':
        case 'b':
        case 'c':
            System.out.println("Value is: " + new String(Character.toChars(a)));
        break;
        default:
            throw new IllegalStateException("This should never happen.");
    }
}

从Java 7开始,你甚至可以用String做同样的事情:

// enum A { BACKWARDS, FOREWARDS, STANDING }
static final Set<String> enumA = new LinkedHashSet<>(Arrays.asList(new String[] {
        "BACKWARDS", "FOREWARDS", "STANDING" }));

// enum B extends A { JUMP }
static final Set<String> enumB = new LinkedHashSet<>(enumA);
static {
    enumB.add("JUMP");
}

使用enum替换:

for (String a : enumB) {
    switch (a) {
        case "BACKWARDS":
        case "FOREWARDS":
        case "STANDING":
            System.out.println("Value is: " + a);
        break;
        default:
            throw new IllegalStateException("This should never happen.");
    }
}

这就是我如何增强枚举继承模式运行时检查在静态初始化器。 BaseKind的#checkEnumExtender检查“extends”enum是否以完全相同的方式声明了基enum的所有值,以便#name()和#ordinal()保持完全兼容。

声明值仍然涉及复制粘贴,但如果有人在基类中添加或修改值而没有更新扩展值,则程序很快就会失败。

不同枚举相互扩展的常见行为:

public interface Kind {
  /**
   * Let's say we want some additional member.
   */
  String description() ;

  /**
   * Standard {@code Enum} method.
   */
  String name() ;

  /**
   * Standard {@code Enum} method.
   */
  int ordinal() ;
}

基准enum,带有验证方法:

public enum BaseKind implements Kind {

  FIRST( "First" ),
  SECOND( "Second" ),

  ;

  private final String description ;

  public String description() {
    return description ;
  }

  private BaseKind( final String description ) {
    this.description = description ;
  }

  public static void checkEnumExtender(
      final Kind[] baseValues,
      final Kind[] extendingValues
  ) {
    if( extendingValues.length < baseValues.length ) {
      throw new IncorrectExtensionError( "Only " + extendingValues.length + " values against "
          + baseValues.length + " base values" ) ;
    }
    for( int i = 0 ; i < baseValues.length ; i ++ ) {
      final Kind baseValue = baseValues[ i ] ;
      final Kind extendingValue = extendingValues[ i ] ;
      if( baseValue.ordinal() != extendingValue.ordinal() ) {
        throw new IncorrectExtensionError( "Base ordinal " + baseValue.ordinal()
            + " doesn't match with " + extendingValue.ordinal() ) ;
      }
      if( ! baseValue.name().equals( extendingValue.name() ) ) {
        throw new IncorrectExtensionError( "Base name[ " + i + "] " + baseValue.name()
            + " doesn't match with " + extendingValue.name() ) ;
      }
      if( ! baseValue.description().equals( extendingValue.description() ) ) {
        throw new IncorrectExtensionError( "Description[ " + i + "] " + baseValue.description()
            + " doesn't match with " + extendingValue.description() ) ;
      }
    }
  }


  public static class IncorrectExtensionError extends Error {
    public IncorrectExtensionError( final String s ) {
      super( s ) ;
    }
  }

}

扩展示例:

public enum ExtendingKind implements Kind {
  FIRST( BaseKind.FIRST ),
  SECOND( BaseKind.SECOND ),
  THIRD( "Third" ),
  ;

  private final String description ;

  public String description() {
    return description ;
  }

  ExtendingKind( final BaseKind baseKind ) {
    this.description = baseKind.description() ;
  }

  ExtendingKind( final String description ) {
    this.description = description ;
  }

}

枚举表示可能值的完整枚举。所以(毫无帮助的)答案是否定的。

举一个实际问题的例子,工作日,周末,工会,一周的天数。我们可以在“天-周”中定义所有的天,但这样就不能表示“工作日”和“周末-天”的特殊属性。

我们可以做的是,有三种枚举类型,并在weekdays/weekend-days和days-of-week之间进行映射。

public enum Weekday {
    MON, TUE, WED, THU, FRI;
    public DayOfWeek toDayOfWeek() { ... }
}
public enum WeekendDay {
    SAT, SUN;
    public DayOfWeek toDayOfWeek() { ... }
}
public enum DayOfWeek {
    MON, TUE, WED, THU, FRI, SAT, SUN;
}

或者,我们可以为day-of-week提供一个开放式接口:

interface Day {
    ...
}
public enum Weekday implements Day {
    MON, TUE, WED, THU, FRI;
}
public enum WeekendDay implements Day {
    SAT, SUN;
}

或者我们可以结合这两种方法:

interface Day {
    ...
}
public enum Weekday implements Day {
    MON, TUE, WED, THU, FRI;
    public DayOfWeek toDayOfWeek() { ... }
}
public enum WeekendDay implements Day {
    SAT, SUN;
    public DayOfWeek toDayOfWeek() { ... }
}
public enum DayOfWeek {
    MON, TUE, WED, THU, FRI, SAT, SUN;
    public Day toDay() { ... }
}