Java - 将Big-Endian转换为Little-Endian [英] Java - Convert Big-Endian to Little-Endian

问题描述

我有以下十六进制字符串：

00000000000008a3a41b85b8b29ad444def299fee21793cd8b9e567eab02cd81

$
$ b

81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000（Big
endian）

我认为我必须反转并交换字符串，但是像这样的东西并没有给我正确的结果：

 字符串十六进制=00000000000008a3a41b85b8b29ad444def299fee21793cd8b9e567eab02cd81; 
 hex = new StringBuilder（hex）.reverse（）。toString（）; 
 
 
 
lock $ 
 
 
 
 
 $ b 81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000 
（应该是）
 
 
交换：
 
 
  public static String hexSwap（String origHex）{
 //从十六进制数中创建数字
 BigInteger orig = new BigInteger（origHex，16）; 
 //获取交换字节
 byte [] origBytes = orig.toByteArray（）; 
 int i = 0; 
 while（origBytes [i] == 0）i ++; 
 //交换字节
 byte [] swapBytes = new byte [origBytes.length]; （$ / $ / $; i< origBytes.length; i ++）{
 swapBytes [i] = origBytes [origBytes.length-i-1]; 
} 
 BigInteger swap = new BigInteger（swapBytes）; 
 return swap.toString（10）; 
} 
 
 hex = hexSwap（hex）; 
  
 
 
 
 
 
结果：
 026053973026883595670517176393898043396144045912271014791797784 
（错误） 
 81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000 
（应该是）
 
 
任何人都可以给我一个如何完成这个任务的例子吗？ 
非常感谢你）
解决方案
您需要交换每个对字符，因为你正在颠倒字节的顺序，而不是nybbles。例如： 
 
 
  public static String reverseHex（String originalHex）{
 // TODO：验证长度是甚至
 int lengthInBytes = originalHex.length（）/ 2; 
 char [] chars = new char [lengthInBytes * 2]; 
 for（int index = 0; index< lengthInBytes; index ++）{
 int reversedIndex = lengthInBytes  -  1  -  index; 
 chars [reversedIndex * 2] = originalHex.charAt（index * 2）; 
 chars [reversedIndex * 2 + 1] = originalHex.charAt（index * 2 + 1）; 
} 
返回新的String（字符）; 
} 
  
 
I have the following hex string:

  
00000000000008a3a41b85b8b29ad444def299fee21793cd8b9e567eab02cd81
but I want it to look like this:

  
81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000 (Big
  endian)
I think I have to reverse and swap the string, but something like this doesn't give me right result:
  String hex = "00000000000008a3a41b85b8b29ad444def299fee21793cd8b9e567eab02cd81";
    hex = new StringBuilder(hex).reverse().toString();



  
Result:
  81dc20bae765e9b8dc39712eef992fed444da92b8b58b14a3a80000000000000
  (wrong)
  81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000
  (should be)
The swapping:
    public static String hexSwap(String origHex) {
        // make a number from the hex
        BigInteger orig = new BigInteger(origHex,16);
        // get the bytes to swap
        byte[] origBytes = orig.toByteArray();
        int i = 0;
        while(origBytes[i] == 0) i++;
        // swap the bytes
        byte[] swapBytes = new byte[origBytes.length];
        for(/**/; i < origBytes.length; i++) {
            swapBytes[i] = origBytes[origBytes.length - i - 1];
        }
        BigInteger swap = new BigInteger(swapBytes);
        return swap.toString(10);
    }

hex = hexSwap(hex);



  
Result: 
  026053973026883595670517176393898043396144045912271014791797784
  (wrong)
  81cd02ab7e569e8bcd9317e2fe99f2de44d49ab2b8851ba4a308000000000000
  (should be)
Can anyone give me a example of how to accomplish this?
Thank you a lot :)
解决方案
You need to swap each pair of characters, as you're reversing the order of the bytes, not the nybbles. So something like:
public static String reverseHex(String originalHex) {
    // TODO: Validation that the length is even
    int lengthInBytes = originalHex.length() / 2;
    char[] chars = new char[lengthInBytes * 2];
    for (int index = 0; index < lengthInBytes; index++) {
        int reversedIndex = lengthInBytes - 1 - index;
        chars[reversedIndex * 2] = originalHex.charAt(index * 2);
        chars[reversedIndex * 2 + 1] = originalHex.charAt(index * 2 + 1);
    }
    return new String(chars);
}


                        
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