在Java中从short转换为byte,反之亦然

Question

我正在尝试将短路转换为2个字节...然后从这2个字节尝试获得相同的短路值.为此,我写了这段代码:

        short oldshort = 700;

        byte 333= (byte) (oldshort);
        byte byte2= (byte) ((oldshort >> 8) & 0xff);

        short newshort = (short) ((byte2 << 8) + byte1);

            System.out.println(oldshort);
        System.out.println(newshort);

对于700(oldshort)的值,newhosrt是444.经过一些测试后,它看起来像这样的代码只适用于某些值.就像...如果oldshort = 50,那么它将正常工作..但如果它是-200,或更大的值超过127(我认为)它不起作用.我想有签名字节,二进制补码等问题...但我无法弄清楚如何解决它.

任何的想法？？在java中以任何本地方式执行此操作？提前致谢!

Answer 1

重新组合时,需要屏蔽byte1以阻止它进行符号扩展.

例如

    short oldshort = 700;

    byte byte1= (byte) (oldshort);
    byte byte2= (byte) ((oldshort >> 8) & 0xff);

    short newshort = (short) ((byte2 << 8) + (byte1&0xFF);

        System.out.println(oldshort);
    System.out.println(newshort);

编辑:java中字节和短路的所有操作实际上都是以整数形式完成的.因此,当您编写时 +byte1,实际发生的是该字节首先被转换为整数(符号扩展).它仍然具有相同的值,但现在有更多的位.然后,我们可以屏蔽底部的8位,从短路中获取原始的8位 - 没有符号.

E.g. short =511 = 0x01FE
     // lots of 0x000's because the operations are done on 32-bit int's
     byte1 = (0x000001FE & 0x000000FF) = (0x01FE & 0xFF) = 0xFE = (byte)-2
     byte2 = 0x1

     newShort = (byte2 << 8) + (byte1 & 0xFF)
              = (0x1 << 8) + (0xFE & 0xFF)
            // since the ops are performed as int's
              = (0x00000001 << 8) + (0xFFFFFFFE & 0x000000FF)
            // 0xFFFFFFFE = -2 
              = (0x00000100) + (0x000000FE)
              = 0x000001FE
              = 511