java 在java中创建一个权限位掩码
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create a permission bit mask in java
提问by mrblah
I want to do something like this:
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
}
Can you do this in Java like that? (I'm coming from a c# background)
你能用 Java 那样做吗？（我来自 ac# 背景）
回答by Chandra Patni
You can easily do it using EnumSet
您可以轻松地使用 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
}
}
回答by Aidamina
This is another option, which is similar to the ordinal solution, except that you can use the | and & operators with this:
这是另一个选项，类似于序数解，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
}
}
or:
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
}
}
回答by MeBigFatGuy
While i wouldn't recommend it, you can ask for the ordinal() of an enum and use that for bit operations. Of course since you can't define what the ordinal is for an enum, you have to insert bogus values to get the ordinals right
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以使序数正确
enum Example {
Bogus, --> 0
This, --> 1
That, --> 2
ThisOrThat --> 3
};
Notice a Bogus enum needed to be introduced so that
注意需要引入一个假枚举，以便
ThisOrThat.ordinal() == This.ordinal() | That.ordinal()
回答by Roman M
If you stuck in the pre Java 7 Era (Android) you can try the following code:
如果您停留在前 Java 7 时代（Android），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());
}
}
}
回答by Upul Bandara
As far as I know bitwise operator is undefined for enum types
据我所知，枚举类型的按位运算符未定义
