java int转byte和long转byte

Posted 2020-08-13

在网络编程中，出于节约带宽或者编码的需要，通常需要以原生方式处理long和int，而不是转换为string。

public class ByteOrderUtils {

public static byte[] int2byte(int res) {
byte[] targets = new byte[4];

targets[3] = (byte) (res & 0xff);// 最低位
targets[2] = (byte) ((res >> 8) & 0xff);// 次低位
targets[1] = (byte) ((res >> 16) & 0xff);// 次高位
targets[0] = (byte) (res >>> 24);// 最高位,无符号右移。
return targets;
}

public static int byteArrayToInt(byte[] b){
byte[] a = new byte[4];
int i = a.length - 1,j = b.length - 1;
for (; i >= 0 ; i--,j--) {//从b的尾部(即int值的低位)开始copy数据
if(j >= 0)
a[i] = b[j];
else
a[i] = 0;//如果b.length不足4,则将高位补0
}
int v0 = (a[0] & 0xff) << 24;//&0xff将byte值无差异转成int,避免Java自动类型提升后,会保留高位的符号位
int v1 = (a[1] & 0xff) << 16;
int v2 = (a[2] & 0xff) << 8;
int v3 = (a[3] & 0xff) ;
return v0 + v1 + v2 + v3;
}

public static byte[] long2byte(long res) {
byte[] buffer = new byte[8];
for (int i = 0; i < 8; i++) {
int offset = 64 - (i + 1) * 8;
buffer[i] = (byte) ((res >> offset) & 0xff);
}
return buffer;
}

public static long byteArrayToLong(byte[] b){
long values = 0;
for (int i = 0; i < 8; i++) {
values <<= 8; values|= (b[i] & 0xff);
}
return values;
}

}