查找使用过的 CRC-16 算法

Question

我正在努力对与 CRC-16 校验和相关的一段数据进行逆向工程。我知道用于计算原始校验和的多项式是 0x8408 但没有别的，我不知道初始值（如果有），最终 XOR 值（如果有），如果输入或结果被反映.. .

似乎有一个已知的 CRC-16 生成器使用事物多项式，CRC-16-CCITT 但尽管我已经尝试了一切，但我还是无法理解正在计算原始校验和。

这是我得到的数据及其各自的校验和。我还在数据和校验和之间包含了一个字节，它是递增的，我不知道它是否经过计算。（看最后两行，数据几乎一样，增量不一样，但校验和是一样的）

|                           DATA                           |Inc|CRC|
|----------------------------------------------------------|---|---|
00 00 00 00 00 00 01 ef f7 fe ef ff fd ef fb fa fd a2 aa 21 01 f4 e0
00 00 00 00 00 00 01 ef f7 fd ef ff fd fe fb fa fd a2 aa 21 02 f4 d1
00 00 00 00 00 00 01 f7 fe fd fd ff fd df ff fb fd a2 aa 21 03 f4 cd
00 00 00 00 00 00 01 f7 fe fe fd ff f7 ef ff fa fd a2 aa 21 04 f4 c2
00 00 00 00 00 00 01 ef f7 fe ef ff fe ef fb fa fd a2 aa 21 05 f4 db
00 00 00 00 00 00 01 ef f7 fe ef ff fd ef fb fa fd a2 aa 21 06 f4 db

Answer 1

每行的最后一个字节似乎是 0xF3 + 除最后一个字节（包括 0xF4）之外的所有字节的负和。此代码适用于 5 个示例：

typedef unsigned char uint8_t;

static uint8_t data0[] = 
   {0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xef,0xf7,0xfe,0xef,
    0xff,0xfd,0xef,0xfb,0xfa,0xfd,0xa2,0xaa,0x21,0x01,0xf4,0xe0};
static uint8_t data1[] = 
   {0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xef,0xf7,0xfd,0xef,
    0xff,0xfd,0xfe,0xfb,0xfa,0xfd,0xa2,0xaa,0x21,0x02,0xf4,0xd1};
static uint8_t data2[] = 
   {0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xf7,0xfe,0xfd,0xfd,
    0xff,0xfd,0xdf,0xff,0xfb,0xfd,0xa2,0xaa,0x21,0x03,0xf4,0xcd};
static uint8_t data3[] = 
   {0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xf7,0xfe,0xfe,0xfd,
    0xff,0xf7,0xef,0xff,0xfa,0xfd,0xa2,0xaa,0x21,0x04,0xf4,0xc2};
static uint8_t data4[] = 
   {0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xef,0xf7,0xfe,0xef,0xff,
    0xfe,0xef,0xfb,0xfa,0xfd,0xa2,0xaa,0x21,0x05,0xf4,0xdb};
static uint8_t data5[] = 
   {0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xef,0xf7,0xfe,0xef,0xff,
    0xfd,0xef,0xfb,0xfa,0xfd,0xa2,0xaa,0x21,0x06,0xf4,0xdb};

int main()
{
size_t i;
uint8_t c = 0xf3;
uint8_t s;
    s = c;
    for(i = 0; i < sizeof(data0)-1; i++)
        s -= data0[i];
    if (data0[i] != s)
        printf("mismatch\n");
    s = c;
    for (i = 0; i < sizeof(data1) - 1; i++)
        s -= data1[i];
    if (data1[i] != s)
        printf("mismatch\n");
    s = c;
    for (i = 0; i < sizeof(data2) - 1; i++)
        s -= data2[i];
    if (data2[i] != s)
        printf("mismatch\n");
    s = c;
    for (i = 0; i < sizeof(data3) - 1; i++)
        s -= data3[i];
    if (data3[i] != s)
        printf("mismatch\n");
    s = c;
    for (i = 0; i < sizeof(data2) - 1; i++)
        s -= data4[i];
    if (data4[i] != s)
        printf("mismatch\n");
    return 0;
}

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