逆向工程 - 德尔福 mt38 CRC (Tricore 1766 Arch) - 吾爱随笔录

我想为 Delphi mt38 计算 CRC32 并找到以下汇编算法并将其转换为 C。似乎该算法是为 n = 256 或 1024 定义的。我想为 n = 1638400 计算 CRC32 并期望得到 CRC32 = 0xFB6568AF对于输入数据 = {0}，但它没有给出预期的答案。我该如何计算？先感谢您。

组装算法：

CheckSum:                               ; CODE XREF: sub_CDBA6
            movh.a          a15, #@HIS(CKS_loop_count)
            lea             a15, [a15]@LOS(CKS_loop_count)
            ld.b            d15, [a15]0
            jz16            d15, loc_CE256
            mov32           d14, #0x100
            j16             loc_CE25A
; ---------------------------------------------------------------------------

loc_CE256:                              ; CODE XREF: CheckSum
            mov32           d14, #0x400

loc_CE25A:                              ; CODE XREF: CheckSum
            mov16           d13, #0
            movh.a          a14, #@HIS(table_address)
            lea             a14, [a14]@LOS(table_address)
            j16             loc_CE2FA
; ---------------------------------------------------------------------------

loc_CE266:                              ; CODE XREF: CheckSum:loc_CE2FA
            movh.a          a15, #@HIS(DATA_address)
            ld32.a          a7, [a15]@LOS(DATA_address)
            ld16.bu         d10, [a7]
            ld32.w          d12, dword_D0000D34
            ld32.w          d11, dword_D0000D30
            jlt.u           d12, d11, loc_CE286
            call32          sub_7FCC0
            add16           d11, d2
            jlt.u           d12, d11, loc_CE2A8

loc_CE286:                              ; CODE XREF: CheckSum+36
            mov.u           d6, #0xFFFF
            ld32.w          d15, CRC_value
            sh32            d7, d15, #-8
            addih           d6, d6, #0xFF
            and16           d7, d6
            xor16           d15, d10
            and32           d15, d15, #0xFF
            addsc16.a       a15, a14, d15, #2
            ld16.w          d15, [a15]
            xor16           d15, d7
            st32.w          CRC_value, d15
loc_CE2A8:                              ; CODE XREF: CheckSum+40
            movh.a          a15, #@HIS(DATA_address)
            ld32.a          a7, [a15]@LOS(DATA_address)
            add16.a         a7, #1
            lea             a15, [a15]@LOS(DATA_address)
            st16.a          [a15], a7
            ld32.w          d15, dword_D0000D34
            add16           d15, #1
            st32.w          dword_D0000D34, d15
            ld32.w          d7, dword_D0000D26
            jlt.u           d15, d7, loc_CE2F8
            ld32.w          d15, CRC_value
            movh.a          a15, #@HIS(CRC_Variable)
            st32.w          [a15]@LOS(CRC_Variable), d15
            mov16           d15, #1
            st32.b          take_CKS_var, d15
            movh.a          a15, #@HIS(CKS_loop_count)
            lea             a15, [a15]@LOS(CKS_loop_count)
            ld.b            d15, [a15]0
            jnz16           d15, loc_CE2FE
            mov16           d15, #1
            movh.a          a15, #@HIS(CKS_loop_count)
            lea             a15, [a15]@LOS(CKS_loop_count)
            st16.b          [a15], d15
            ret16
; ---------------------------------------------------------------------------

loc_CE2F8:                              ; CODE XREF: CheckSum+84
            add16           d13, #1

loc_CE2FA:                              ; CODE XREF: CheckSum+22
            jlt.u           d13, d14, loc_CE266

loc_CE2FE:                              ; CODE XREF: CheckSum+A6
            ret16
; End of function CheckSum

转换后的C算法：

uint32_t crc32(uint32_t n, uint32_t *data) {
    uint32_t i = 0;
    uint32_t ch;
    crc = 0xFFFFFFFF;
    while (i < n) {
        ch = data[i];
        crc = ((crc >> 8) & 0xFFFFFF) ^
              *(crc32_table + ((((crc ^ ch) & 0xFF) << 2) & 0xFF));
        i++;
    }
    return crc;
}