Java 将 32 位无符号整数（大端）转换为 long 和 back
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Converting 32-bit unsigned integer (big endian) to long and back
提问by Aviram
I have a byte[4] which contains a 32-bit unsigned integer (in big endian order) and I need to convert it to long (as int can't hold an unsigned number).
我有一个 byte[4]，它包含一个 32 位无符号整数（按大端顺序），我需要将其转换为 long（因为 int 不能保存无符号数）。
Also, how do I do it vice-versa (i.e. from long that contains a 32-bit unsigned integer to byte[4])?
另外，反之亦然，我该怎么做（即从包含 32 位无符号整数的 long 到 byte[4]）？
采纳答案by Edwin Dalorzo
Sounds like a work for the ByteBuffer.
听起来像是ByteBuffer的作品。
Somewhat like
有点像
public static void main(String[] args) {
byte[] payload = toArray(-1991249);
int number = fromArray(payload);
System.out.println(number);
}
public static int fromArray(byte[] payload){
ByteBuffer buffer = ByteBuffer.wrap(payload);
buffer.order(ByteOrder.BIG_ENDIAN);
return buffer.getInt();
}
public static byte[] toArray(int value){
ByteBuffer buffer = ByteBuffer.allocate(4);
buffer.order(ByteOrder.BIG_ENDIAN);
buffer.putInt(value);
buffer.flip();
return buffer.array();
}
回答by Hot Licks
You can use ByteBuffer, or you can do it the old-fashioned way:
您可以使用 ByteBuffer，也可以使用老式的方法：
long result = 0x00FF & byteData[0];
result <<= 8;
result += 0x00FF & byteData[1];
result <<= 8;
result += 0x00FF & byteData[2];
result <<= 8;
result += 0x00FF & byteData[3];
回答by Sam Barnum
Guava has useful classes for dealing with unsigned numeric values.
Guava 具有处理无符号数值的有用类。
