在小端以短值转换字节数组值 [英] Converting byte array values in little endian order to short values
问题描述
我有一个字节数组,其中数组中的数据实际上是短期的数据。该字节以Little Endian下令:
I have a byte array where the data in the array is actually short data. The bytes are ordered in little endian:
3,1,-48,0,-15,0,36,1
3, 1, -48, 0, -15, 0, 36, 1
,当在转换为短值的结果:
Which when converted to short values results in:
259,208,241,292
259, 208, 241, 292
有在Java中的简单方法的字节值转换为它们的对应的短的值?我可以写一个循环,只需要每一个高字节,由8位,其低字节接班和OR,但是,有一个性能命中。
Is there a simple way in Java to convert the byte values to their corresponding short values? I can write a loop that just takes every high byte and shift it by 8 bits and OR it with its low byte, but that has a performance hit.
推荐答案
通过 java.nio.ByteBuffer中你可以指定你想要的字节顺序:<一href=\"http://docs.oracle.com/javase/7/docs/api/java/nio/ByteBuffer.html#order%28java.nio.ByteOrder%29\">order().
With java.nio.ByteBuffer you may specify the endianness you want: order().
的ByteBuffer有方法提取数据字节,CHAR,<一个href=\"http://docs.oracle.com/javase/7/docs/api/java/nio/ByteBuffer.html#getShort%28%29\">getShort(), 调用getInt(),长一倍。 ..
ByteBuffer have methods to extract data as byte, char, getShort(), getInt(), long, double...
下面是一个例子如何使用它:
Here's an example how to use it:
ByteBuffer bb = ByteBuffer.wrap(byteArray);
bb.order( ByteOrder.LITTLE_ENDIAN);
while( bb.hasRemaining()) {
short v = bb.getShort();
/* Do something with v... */
}
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