将java.net.InetAddress转换为long

Question

我想转换一个java.net.InetAddress和我签署/签署的问题.这样的痛苦.

我在Java中读取从短到字节的转换,反之亦然,为什么字节b =(字节)0xFF等于整数-1？

结果得出:

     final byte [] pumpeIPAddressRaw =
        java.net.InetAddress.getByName (pumpeIPAddressName).getAddress ();

     final long pumpeIPAddress =
         ((pumpeIPAddressRaw [0] & 0xFF) << (3*8)) +
         ((pumpeIPAddressRaw [1] & 0xFF) << (2*8)) +
         ((pumpeIPAddressRaw [2] & 0xFF) << (1*8)) +
         (pumpeIPAddressRaw [3] &  0xFF);

     android.util.Log.i (
        Application.TAG, "LOG00120: Setzte Pumpen Addresse : " +
        pumpeIPAddress + ":" + pumpeIPPort);

并猜猜日志仍然显示:

04-10 13:12:07.398 I/ch.XXXX.remote.Application(24452): LOG00120: Setzte Pumpen Addresse : -1063035647:27015

有人知道我还在做错吗？

Answer 1

& 0xff块转换过程中从符号扩展byte到int,但你的表达也包含从转换int到long你需要阻止这个转换过程中符号扩展,以及:

final long pumpeIPAddress =
      (((pumpeIPAddressRaw [0] & 0xFF) << (3*8)) + 
      ((pumpeIPAddressRaw [1] & 0xFF) << (2*8)) +
      ((pumpeIPAddressRaw [2] & 0xFF) << (1*8)) +
      (pumpeIPAddressRaw [3] &  0xFF)) & 0xffffffffl; 

或者,可以从转换byte到long在一个单一的步骤中,通过标记的第二操作数& 0xff的操作,long使用l后缀:

final long pumpeIPAddress =
      ((pumpeIPAddressRaw [0] & 0xFFl) << (3*8)) + 
      ((pumpeIPAddressRaw [1] & 0xFFl) << (2*8)) +
      ((pumpeIPAddressRaw [2] & 0xFFl) << (1*8)) +
      (pumpeIPAddressRaw [3] &  0xFFl);