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如何使用按位运算符将多个整数值传递给 Java 函数？

发布于 2024-07-14 15:16:52 字数 1032 浏览 9 评论 0原文

在应用程序框架中，我不断看到允许您将多个 Int 值（通常用于代替枚举）传递到函数中的框架。

例如：

public class Example
{ 
    public class Values
    {
        public static final int ONE = 0x7f020000;
        public static final int TWO = 0x7f020001;
        public static final int THREE = 0x7f020002;
        public static final int FOUR = 0x7f020003;
        public static final int FIVE = 0x7f020004;
    }

    public static void main(String [] args)
    {
        // should evaluate just Values.ONE
        Example.multiValueExample(Values.ONE);

        // should evalueate just values Values.ONE,  Values.THREE, Values.FIVE
        Example.multiValueExample(Values.ONE | Values.THREE | Values.FIVE);

        // should evalueate just values Values.TWO , Values.FIVE
        Example.multiValueExample(Values.TWO | Values.FIVE);
    }

    public static void multiValueExample(int values){
        // Logic that properly evaluates bitwise values
        ...
    }
}

那么 multiValueExample 中应该存在什么逻辑才能正确评估使用按位运算符传入的多个 int 值？

原文

In application frameworks I keep seeing frameworks that allow you to pass in multiple Int values (generally used in place of an enum) into a function.

For example:

public class Example
{ 
    public class Values
    {
        public static final int ONE = 0x7f020000;
        public static final int TWO = 0x7f020001;
        public static final int THREE = 0x7f020002;
        public static final int FOUR = 0x7f020003;
        public static final int FIVE = 0x7f020004;
    }

    public static void main(String [] args)
    {
        // should evaluate just Values.ONE
        Example.multiValueExample(Values.ONE);

        // should evalueate just values Values.ONE,  Values.THREE, Values.FIVE
        Example.multiValueExample(Values.ONE | Values.THREE | Values.FIVE);

        // should evalueate just values Values.TWO , Values.FIVE
        Example.multiValueExample(Values.TWO | Values.FIVE);
    }

    public static void multiValueExample(int values){
        // Logic that properly evaluates bitwise values
        ...
    }
}

So what logic should exist in multiValueExample for me to properly evaluate multiple int values being passed in using the bitwise operator?

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评论（8）

影子的影子 2024-07-21 15:16:52

您的值应该是 2 的幂。

这样，当您对它们进行按位或运算时，您就不会丢失任何信息。

public static final int ONE   = 0x01;
public static final int TWO   = 0x02;
public static final int THREE = 0x04;
public static final int FOUR  = 0x08;
public static final int FIVE  = 0x10;

等等

然后你可以这样做：

public static void main(String [] args) {
    Example.multiValueExample(Values.ONE | Values.THREE | Values.FIVE);
}

public static void multiValueExample(int values){
    if ((values & Values.ONE) == Values.ONE) {
    }

    if ((values & Values.TWO) == Values.TWO) {
    }

    // etc.
}

Your values should be powers of 2.

That way, you don't lose any information when you bitwise-OR them.

public static final int ONE   = 0x01;
public static final int TWO   = 0x02;
public static final int THREE = 0x04;
public static final int FOUR  = 0x08;
public static final int FIVE  = 0x10;

etc.

Then you can do this:

public static void main(String [] args) {
    Example.multiValueExample(Values.ONE | Values.THREE | Values.FIVE);
}

public static void multiValueExample(int values){
    if ((values & Values.ONE) == Values.ONE) {
    }

    if ((values & Values.TWO) == Values.TWO) {
    }

    // etc.
}
回复 原文
天荒地未老 2024-07-21 15:16:52

正如已经提到的，考虑使用枚举而不是位值。

根据 Effective Java 2：“第 32 项：使用 EnumSet 而不是位字段"

使用 EnumSet 对于内存使用也很方便。

这是一个例子：

package enums;

import java.util.EnumSet;
import java.util.Set;

public class Example {
  public enum Values {
    ONE, TWO, THREE, FOUR, FIVE
  }

  public static void main(String[] args) {
    // should evaluate just Values.ONE
    Example.multiValueExample(EnumSet.of(Values.ONE));

    // should evalueate just values Values.ONE, Values.THREE, Values.FIVE
    Example.multiValueExample(EnumSet.of(Values.ONE, Values.THREE, Values.FIVE));

    // should evalueate just values Values.TWO , Values.FIVE
    Example.multiValueExample(EnumSet.of(Values.TWO, Values.FIVE));
  }

  public static void multiValueExample(Set<Values> values) {
    if (values.contains(Values.ONE)) {
      System.out.println("One");
    }

    // Other checks here...

    if (values.contains(Values.FIVE)) {
      System.out.println("Five");
    }
  }
}

As was already mentioned, consider use of enums instead of bit values.

According to Effective Java 2: "Item 32: Use EnumSet instead of bit fields"

Usage of EnumSet is quite effective for memory usage and very convenient.

Here is an example:

package enums;

import java.util.EnumSet;
import java.util.Set;

public class Example {
  public enum Values {
    ONE, TWO, THREE, FOUR, FIVE
  }

  public static void main(String[] args) {
    // should evaluate just Values.ONE
    Example.multiValueExample(EnumSet.of(Values.ONE));

    // should evalueate just values Values.ONE, Values.THREE, Values.FIVE
    Example.multiValueExample(EnumSet.of(Values.ONE, Values.THREE, Values.FIVE));

    // should evalueate just values Values.TWO , Values.FIVE
    Example.multiValueExample(EnumSet.of(Values.TWO, Values.FIVE));
  }

  public static void multiValueExample(Set<Values> values) {
    if (values.contains(Values.ONE)) {
      System.out.println("One");
    }

    // Other checks here...

    if (values.contains(Values.FIVE)) {
      System.out.println("Five");
    }
  }
}
回复 原文
甜点 2024-07-21 15:16:52

您将整数值设置为 2 的幂，以便每个枚举值都是二进制表示形式中的单个位。

int ONE = 0x1;    //0001
int TWO = 0x2;    //0010
int THREE = 0x4;  //0100
int FOUR = 0x8;   //1000

然后使用按位或来组合值，并使用按位与来测试设置值。

int test_value = (ONE | FOUR);   //-> 1001
bool has_one = (test_value & ONE) != 0;  //-> 1001 & 0001 -> 0001 -> true

You setup the integer values to be powers of two so that each enumerated value is a single bit in the binary representation.

int ONE = 0x1;    //0001
int TWO = 0x2;    //0010
int THREE = 0x4;  //0100
int FOUR = 0x8;   //1000

Then you use bit-wise OR to combine values and bitwise AND to test set values.

int test_value = (ONE | FOUR);   //-> 1001
bool has_one = (test_value & ONE) != 0;  //-> 1001 & 0001 -> 0001 -> true
回复 原文
七分※倦醒 2024-07-21 15:16:52

您与 | 结合的值 （二进制 OR，不是逻辑 OR [即 ||]）在其位表示中不得有重叠的“1”。 例如，

ONE = 0x1 =   0000 0001
TWO = 0x2 =   0000 0010
THREE = 0x3 = 0000 0011
FOUR = 0x4 =  0000 0100

那么你可以将 ONE 和 TWO 结合起来，例如：

ONE | TWO = 0000 0011

但你无法区分 ONE | 三中二，因为存在重叠位。 因此，您组合的数字应该是 2 的幂，这样它们在进行“或”运算时不会重叠。 要测试数字是否在“值”中传递，请执行以下操作：

if (values & ONE) {
    // ... then ONE was set
}

为了更好地理解其工作原理和原理，我建议您阅读一些有关二进制表示和逻辑的内容。 一个好地方是《汇编艺术》第 3 章。

The values you combine with | (binary OR, not logical OR [which is ||]) must not have overlapping "1"s in their bit representation. For example,

ONE = 0x1 =   0000 0001
TWO = 0x2 =   0000 0010
THREE = 0x3 = 0000 0011
FOUR = 0x4 =  0000 0100

Then you can combine ONE and TWO, for example:

ONE | TWO = 0000 0011

But you can't distinguish ONE | TWO from THREE, because there are overlapping bits. The numbers you combine should thus be powers of two, such that they don't overlap when OR'ed together. To test if a number was passed in "values", do:

if (values & ONE) {
    // ... then ONE was set
}

To better understand why and how this works, I recommend you read a bit on binary representation and logic. A good place is Chapter 3 of the Art of Assembly.

回复 原文
叫嚣ゝ 2024-07-21 15:16:52

如果它们是 2 的幂，您将执行类似于下面代码中的“display”方法的操作。

这里是维基百科中关于该主题的链接，它应该解释为什么你想要 2 的幂。

public class Main
{
    private static final int A = 0x01;
    private static final int B = 0x02;
    private static final int C = 0x04;

    public static void main(final String[] argv)
    {
        display(A);
        display(B);
        display(C);
        display(A | A);
        display(A | B);
        display(A | C);
        display(B | A);
        display(B | B);
        display(B | C);
        display(C | A);
        display(C | B);
        display(C | C);
        display(A | A | A);
        display(A | A | B);
        display(A | A | C);
        display(A | B | A);
        display(A | B | B);
        display(A | B | C);
        display(A | C | A);
        display(A | C | B);
        display(A | C | C);
        display(B | A | A);
        display(B | A | B);
        display(B | A | C);
        display(B | B | A);
        display(B | B | B);
        display(B | B | C);
        display(B | C | A);
        display(B | C | B);
        display(B | C | C);
        display(C | A | A);
        display(C | A | B);
        display(C | A | C);
        display(C | B | A);
        display(C | B | B);
        display(C | B | C);
        display(C | C | A);
        display(C | C | B);
        display(C | C | C);
    }

    private static void display(final int val)
    {
        if((val & A) != 0)
        {
            System.out.print("A");
        }

        if((val & B) != 0)
        {
            System.out.print("B");
        }

        if((val & C) != 0)
        {
            System.out.print("C");
        }

        System.out.println();
    }
}

Well if they are powers of 2 you would do something like the "display" method in the code below.

Here is a link in wikipedia on the topic as well which should explain why you want powers of 2.

public class Main
{
    private static final int A = 0x01;
    private static final int B = 0x02;
    private static final int C = 0x04;

    public static void main(final String[] argv)
    {
        display(A);
        display(B);
        display(C);
        display(A | A);
        display(A | B);
        display(A | C);
        display(B | A);
        display(B | B);
        display(B | C);
        display(C | A);
        display(C | B);
        display(C | C);
        display(A | A | A);
        display(A | A | B);
        display(A | A | C);
        display(A | B | A);
        display(A | B | B);
        display(A | B | C);
        display(A | C | A);
        display(A | C | B);
        display(A | C | C);
        display(B | A | A);
        display(B | A | B);
        display(B | A | C);
        display(B | B | A);
        display(B | B | B);
        display(B | B | C);
        display(B | C | A);
        display(B | C | B);
        display(B | C | C);
        display(C | A | A);
        display(C | A | B);
        display(C | A | C);
        display(C | B | A);
        display(C | B | B);
        display(C | B | C);
        display(C | C | A);
        display(C | C | B);
        display(C | C | C);
    }

    private static void display(final int val)
    {
        if((val & A) != 0)
        {
            System.out.print("A");
        }

        if((val & B) != 0)
        {
            System.out.print("B");
        }

        if((val & C) != 0)
        {
            System.out.print("C");
        }

        System.out.println();
    }
}
回复 原文
鲸落 2024-07-21 15:16:52

当每一位都很重要时，使用位掩码很受欢迎。 今天实现此目的的另一种方法是使用更易于操作和扩展的枚举。

import static Example.Values.*;
import java.util.Arrays;

public class Example {
    public enum Values { ONE, TWO, THREE, FOUR, FIVE }

    public static void main(String [] args) {
        // should evaluate just Values.ONE
        multiValueExample(ONE);

        // should evaluate just values Values.ONE,  Values.THREE, Values.FIVE
        multiValueExample(ONE, THREE, FIVE);

        // should evaluate just values Values.TWO , Values.FIVE
        multiValueExample(TWO, FIVE);
    }

    public static void multiValueExample(Values... values){
        // Logic that properly evaluates
        System.out.println(Arrays.asList(values));
        for (Values value : values) {
            // do something.
        }
    }
}

Using bit masks were popular when every bit counted. Another way to do this today is to use enums with are simpler to manipulate and extend.

import static Example.Values.*;
import java.util.Arrays;

public class Example {
    public enum Values { ONE, TWO, THREE, FOUR, FIVE }

    public static void main(String [] args) {
        // should evaluate just Values.ONE
        multiValueExample(ONE);

        // should evaluate just values Values.ONE,  Values.THREE, Values.FIVE
        multiValueExample(ONE, THREE, FIVE);

        // should evaluate just values Values.TWO , Values.FIVE
        multiValueExample(TWO, FIVE);
    }

    public static void multiValueExample(Values... values){
        // Logic that properly evaluates
        System.out.println(Arrays.asList(values));
        for (Values value : values) {
            // do something.
        }
    }
}
回复 原文
一梦浮鱼 2024-07-21 15:16:52

首先，您不能以这种方式定义值来进行按位比较。 相反，设置不同的位：

public static final int ONE   = 0x1;  // First bit is set
public static final int TWO   = 0x2;  // Second bit is set
public static final int THREE = 0x4;  // Third bit is set
public static final int FOUR  = 0x8;  // Fourth bit is set
public static final int FIVE  = 0x10; // Fifth bit is set

其次，您可能应该使用 java.util.BitSet 用于这些类型的操作：

BitSet bits = new BitSet(5);
bits.set(2);
bits.set(4);

System.out.println("these bits are set: " + bits);
// Prints "these bits are set: {2, 4}"

BitSet otherBits = new BitSet(5);
otherBits.set(3);
otherBits.set(4);

System.out.println("these bits are set: " + bits.or(otherBits));
// Prints "these bits are set: {2, 3, 4}"

First, you can't define the Values that way to do bitwise comparisons. Instead, set different bits:

public static final int ONE   = 0x1;  // First bit is set
public static final int TWO   = 0x2;  // Second bit is set
public static final int THREE = 0x4;  // Third bit is set
public static final int FOUR  = 0x8;  // Fourth bit is set
public static final int FIVE  = 0x10; // Fifth bit is set

Second, you should probably be using java.util.BitSet for these sorts of operations:

BitSet bits = new BitSet(5);
bits.set(2);
bits.set(4);

System.out.println("these bits are set: " + bits);
// Prints "these bits are set: {2, 4}"

BitSet otherBits = new BitSet(5);
otherBits.set(3);
otherBits.set(4);

System.out.println("these bits are set: " + bits.or(otherBits));
// Prints "these bits are set: {2, 3, 4}"
回复 原文
情痴 2024-07-21 15:16:52

有关按位运算的 Java 教程章节位于

http://java .sun.com/docs/books/tutorial/java/nutsandbolts/op3.html

非常简洁，但很有参考价值。

The Java Tutorial chapter on bitwise operations are at

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/op3.html

It is very concise but good for reference.

回复 原文
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