将整数转换为字节数组（Java）

Convert integer into byte array (Java)

提问by Buttercup

what's a fast way to convert an Integerinto a Byte Array?

将 an 转换Integer为 a的快速方法是Byte Array什么？

e.g. 0xAABBCCDD => {AA, BB, CC, DD}

例如 0xAABBCCDD => {AA, BB, CC, DD}

采纳答案by Gregory Pakosz

Have a look at the ByteBufferclass.

看看ByteBuffer类。

ByteBuffer b = ByteBuffer.allocate(4);
//b.order(ByteOrder.BIG_ENDIAN); // optional, the initial order of a byte buffer is always BIG_ENDIAN.
b.putInt(0xAABBCCDD);

byte[] result = b.array();

Setting the byte order ensures that result[0] == 0xAA, result[1] == 0xBB, result[2] == 0xCCand result[3] == 0xDD.

设置字节顺序保证了result[0] == 0xAA，result[1] == 0xBB，result[2] == 0xCC和result[3] == 0xDD。

Or alternatively, you could do it manually:

或者，您可以手动执行此操作：

byte[] toBytes(int i)
{
  byte[] result = new byte[4];

  result[0] = (byte) (i >> 24);
  result[1] = (byte) (i >> 16);
  result[2] = (byte) (i >> 8);
  result[3] = (byte) (i /*>> 0*/);

  return result;
}

The ByteBufferclass was designed for such dirty hands tasks though. In fact the private java.nio.Bitsdefines these helper methods that are used by ByteBuffer.putInt():

不过，该ByteBuffer课程是为此类脏手任务而设计的。事实上，私有java.nio.Bits定义了这些被使用的辅助方法ByteBuffer.putInt()：

private static byte int3(int x) { return (byte)(x >> 24); }
private static byte int2(int x) { return (byte)(x >> 16); }
private static byte int1(int x) { return (byte)(x >>  8); }
private static byte int0(int x) { return (byte)(x >>  0); }

回答by notnoop

You can use BigInteger:

您可以使用BigInteger：

From Integers:

从整数：

byte[] array = BigInteger.valueOf(0xAABBCCDD).toByteArray();
System.out.println(Arrays.toString(array))
// --> {-86, -69, -52, -35 }

The returned array is of the size that is needed to represent the number, so it could be of size 1, to represent 1 for example. However, the size cannot be more than four bytes if an int is passed.

返回的数组具有表示数字所需的大小，因此它的大小可能为 1，例如表示 1。但是，如果传递的是 int，则大小不能超过四个字节。

From Strings:

从字符串：

BigInteger v = new BigInteger("AABBCCDD", 16);
byte[] array = v.toByteArray();

However, you will need to watch out, if the first byte is higher 0x7F(as is in this case), where BigInteger would insert a 0x00 byte to the beginning of the array. This is needed to distinguish between positive and negative values.

但是，您需要注意，如果第一个字节更高0x7F（在这种情况下），BigInteger 会在数组的开头插入一个 0x00 字节。这是区分正值和负值所必需的。

回答by Pascal Thivent

Using BigInteger:

使用BigInteger：

private byte[] bigIntToByteArray( final int i ) {
    BigInteger bigInt = BigInteger.valueOf(i);      
    return bigInt.toByteArray();
}

Using DataOutputStream:

使用DataOutputStream：

private byte[] intToByteArray ( final int i ) throws IOException {      
    ByteArrayOutputStream bos = new ByteArrayOutputStream();
    DataOutputStream dos = new DataOutputStream(bos);
    dos.writeInt(i);
    dos.flush();
    return bos.toByteArray();
}

Using ByteBuffer:

使用ByteBuffer：

public byte[] intToBytes( final int i ) {
    ByteBuffer bb = ByteBuffer.allocate(4); 
    bb.putInt(i); 
    return bb.array();
}

回答by daz

use this function it works for me

使用这个功能对我有用

public byte[] toByteArray(int value) {
    return new byte[] {
            (byte)(value >> 24),
            (byte)(value >> 16),
            (byte)(value >> 8),
            (byte)value};
}

it translates the int into a byte value

它将 int 转换为字节值

回答by Pang

If you like Guava, you may use its Intsclass:

如果你喜欢Guava，你可以使用它的Ints类：

---

---

For int→ byte[], use toByteArray():

对于int→ byte[]，使用toByteArray()：

byte[] byteArray = Ints.toByteArray(0xAABBCCDD);

Result is {0xAA, 0xBB, 0xCC, 0xDD}.

结果是{0xAA, 0xBB, 0xCC, 0xDD}。

---

---

Its reverse is fromByteArray()or fromBytes():

它的反面是fromByteArray()或fromBytes()：

int intValue = Ints.fromByteArray(new byte[]{(byte) 0xAA, (byte) 0xBB, (byte) 0xCC, (byte) 0xDD});
int intValue = Ints.fromBytes((byte) 0xAA, (byte) 0xBB, (byte) 0xCC, (byte) 0xDD);

Result is 0xAABBCCDD.

结果是0xAABBCCDD。

回答by user3424779

Here's a method that should do the job just right.

这是一种应该正确完成工作的方法。

public byte[] toByteArray(int value)
{
    final byte[] destination = new byte[Integer.BYTES];
    for(int index = Integer.BYTES - 1; index >= 0; index--)
    {
        destination[i] = (byte) value;
        value = value >> 8;
    };
    return destination;
};

回答by Muskovets

It's my solution:

这是我的解决方案：

public void getBytes(int val) {
    byte[] bytes = new byte[Integer.BYTES];
    for (int i = 0;i < bytes.length; i ++) {
        int j = val % Byte.MAX_VALUE;
        bytes[i] = (j == 0 ? Byte.MAX_VALUE : j);
    }
}

Also Stringy method:

还有Stringy方法：

public void getBytes(int val) {
    String hex = Integer.toHexString(val);
    byte[] val = new byte[hex.length()/2]; // because byte is 2 hex chars
    for (int i = 0; i < hex.length(); i+=2)
        val[i] = Byte.parseByte("0x" + hex.substring(i, i+2), 16);
    return val;
}

回答by wai

This will help you.

这会帮助你。

import java.nio.ByteBuffer;
import java.util.Arrays;

public class MyClass
{
    public static void main(String args[]) {
        byte [] hbhbytes = ByteBuffer.allocate(4).putInt(16666666).array();

        System.out.println(Arrays.toString(hbhbytes));
    }
}

回答by Matt

Can also shift -

还可以移——

byte[] ba = new byte[4];
int val = Integer.MAX_VALUE;

for(byte i=0;i<4;i++)
    ba[i] = (byte)(val >> i*8);
    //ba[3-i] = (byte)(val >> i*8); //Big-endian

回答by helmy

Simple solution which properly handles ByteOrder:

正确处理 ByteOrder 的简单解决方案：

ByteBuffer.allocate(4).order(ByteOrder.nativeOrder()).putInt(yourInt).array();

ByteBuffer.allocate(4).order(ByteOrder.nativeOrder()).putInt(yourInt).array();