java - java中带符号字节的CRC32

Question

我正在尝试将 C CRC32 代码转换为 java。下面的代码是我想出的。但它的工作方式似乎有所不同。用 C 计算的预期 CRC32 是 13 82 D8 46)，但以下代码的输出是“最终的 crc 是 -2084771805，十六进制 ffffffff83bce823”。谁能告诉我为什么？

C代码

uint32 crc32_update(uint32 crc, const uint8_t *data, uint16 data_len)
{

uint16_t tbl_idx;

while (data_len--) {
    tbl_idx = crc ^ (*data >> (0 * 4));
    crc = crc_table[tbl_idx & 0x0f] ^ (crc >> 4);
    tbl_idx = crc ^ (*data >> (1 * 4));
    crc = crc_table[tbl_idx & 0x0f] ^ (crc >> 4);

    data++;
}
return crc & 0xffffffff;
}

JAVA代码。

public class crc32trial_3 {


static  final  long crc_table[] = new long[] {
    0x00000000, 0x1db71064, 0x3b6e20c8, 0x26d930ac,
    0x76dc4190, 0x6b6b51f4, 0x4db26158, 0x5005713c,
    0xedb88320, 0xf00f9344, 0xd6d6a3e8, 0xcb61b38c,
    0x9b64c2b0, 0x86d3d2d4, 0xa00ae278, 0xbdbdf21c
};



public static long crc32_init()
{
    return 0xffffffff;
}

private static  long crc32_update(long crc, byte[] data, int data_len)
 {
     int tbl_idx;

    for(int i = 0 ; i <  data_len ; i ++) {
         tbl_idx = (int)crc ^ (data[i] >> (0 * 4));
         crc = crc_table[tbl_idx & 0x0f] ^ (crc >> 4);
         tbl_idx = (int)crc ^ (data[i] >> (1 * 4));
         crc = crc_table[tbl_idx & 0x0f] ^ (crc >> 4);

         //data++;
     }

     return crc & 0xffffffff;

 }



 public static void main(String args[])
 {
     long intialcrc = crc32_init();
     long crc;


     System.out.println("the intail crc = " + intialcrc);
     byte[] packets = new byte[]{ 88,0,1,0,0,0,0,0,0,0,0,0,-1,-1,-1,-1,1,0,0,0,-1,-1,-1,-1,0,0,-56,-46,-117,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0,-68,118 };

     byte[] totalLenght = new byte[]{100,0,0,0};
     byte[] totalSettingBlock = new byte[]{2};
      crc = crc32_update(intialcrc,totalLenght,4);
      crc = crc32_update(crc, totalSettingBlock,1);
      int temp = 28 + 72;
      crc = crc32_update(crc,packets, temp);
      long finalcrc = crc;




     System.out.println(" the final crc is " + finalcrc + "  in hex " +  Long.toHexString(finalcrc));

 }


}

Answer 1

您需要在所有十六进制常量的末尾添加“L”。然后我得到 46d81382 作为结果，它接近，但你所说的排列是预期的。