17

我想做这样的事情：

public enum Permissions
{
    CanBlah1,
    CanBlah2,
    CanBlah3
}

byte[] userPerm = Permissions.CanBlah1 | Permissions.CanBlah2;

// check permssions
//
if(userPerm && Permissions.CanBlah1 == Permissions.CanBlah1)
{
      // do something
}

你能像这样在Java中做到这一点吗？（我来自ac#背景）

4

5 回答 5

38

您可以使用EnumSet

import java.util.EnumSet;

import static java.util.EnumSet.of;
import static java.util.EnumSet.range;
import static so.User.Permissions.CanBlah1;
import static so.User.Permissions.CanBlah2;
import static so.User.Permissions.CanBlah3;

public class User {
    public enum Permissions {
        CanBlah1,
        CanBlah2,
        CanBlah3
    }

    public static void main(String[] args) throws Exception {
        EnumSet<Permissions> userPerms = of(CanBlah1, CanBlah2);
        System.out.println(userPerms.contains(CanBlah1)); //true
        System.out.println(userPerms.contains(CanBlah2)); //true
        System.out.println(userPerms.contains(CanBlah3)); //false
        System.out.println(userPerms.containsAll(of(CanBlah1, CanBlah3))); //false
        System.out.println(userPerms.containsAll(range(CanBlah1, CanBlah2))); //true
        System.out.println(userPerms.containsAll(range(CanBlah1, CanBlah3))); //false
    }

}
于 2010-01-01T02:42:27.190 回答
7

这是另一个选项，类似于序数解决方案，不同之处在于您可以使用 | 和 & 运算符：

 public enum Permissions {
     CanBlah1(1),
     CanBlah2(2),
     CanBlah3(4);

     public int value;

     Permissions(int value) {
         this.value = value;
     }
     public int value() {
      return value;
     }
 }

 public static void main(String[] args) {  
    int userPerm = Permissions.CanBlah1.value() | Permissions.CanBlah2.value();
    // check permssions
    //
    if((userPerm & Permissions.CanBlah1.value()) == Permissions.CanBlah1.value())
    {
        // do something
    }
 }

或者：

 public enum Permissions {
         CanBlah1,
         CanBlah2,
         CanBlah3;

         public int value() {
            return 1<<ordinal();
         }
     }

     public static void main(String[] args) {  
        int userPerm = Permissions.CanBlah1.value() | Permissions.CanBlah2.value();
        // check permssions
        //
        if((userPerm & Permissions.CanBlah1.value()) == Permissions.CanBlah1.value())
        {
            // do something
        }
     }
于 2010-12-25T18:19:36.063 回答
3

虽然我不推荐它，但您可以要求枚举的 ordinal() 并将其用于位操作。当然，由于您无法定义枚举的序数，因此您必须插入虚假值才能使序数正确

enum Example {
   Bogus,            --> 0
   This,             --> 1
   That,             --> 2
   ThisOrThat        --> 3
};

请注意，需要引入一个伪造的枚举，以便

ThisOrThat.ordinal() == This.ordinal() | That.ordinal()
于 2010-01-03T04:33:49.957 回答
2

如果您停留在 Java 7 Era (Android) 之前，您可以尝试以下代码：

public enum STUFF_TO_BIT_BASK {
THIS,THAT,OTHER;

public static int getBitMask(STUFF_TO_BIT_BASK... masks) {
    int res = 0;

    for (STUFF_TO_BIT_BASK cap : masks) {
        res |= (int) Math.pow(2, cap.ordinal());
    }

    return res;
}

public boolean is(int maskToCheck){
    return maskToCheck | (int) Math.pow(2, this.ordinal());
}

}

于 2014-02-04T15:44:15.287 回答
0

据我所知，位运算符对于枚举类型是未定义的

于 2010-01-01T02:38:20.240 回答