逆向解析單片機串口校驗算法,幀尾FC的前2個字節(jié)是怎么算出來的?

ID:619688 · 發(fā)表于 2021-4-13 15:50

幫忙解析一下串口校驗算法,看看幀尾FC的前2個是怎么算出來的，以下是單片機發(fā)出來的數(shù)據(jù)：
FD 88 01 01 00 00 00 00 00 00 00 8B 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 16 91 FC
FD 88 01 01 04 00 00 00 00 00 00 8B 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 CA 9D FC
FD 88 01 01 08 00 00 00 00 00 00 8B 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 BE A8 FC
FD 88 01 01 0C 00 00 00 00 00 00 8B 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 62 A4 FC
FD 88 01 01 04 00 00 00 00 00 00 88 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 FB 00 BB FC
FD 88 01 01 40 00 00 00 00 00 00 8B 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 07 FB 01 FC
FD 88 01 01 04 00 00 00 00 00 00 80 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 7C AB FC

ID:247164 · 發(fā)表于 2021-4-13 15:50

夢想未央發(fā)表于 2021-4-14 09:23
計算參考：【CRC筆記】CRC-16 XMODEM C語言實現(xiàn) https://blog.csdn.net/u012028275/article/details/11206 ...

采用的是CRC-16/XMODEM校驗方式，F(xiàn)D為幀頭，不參于校驗，數(shù)據(jù)位從88開始，F(xiàn)C為幀尾，前兩位為校驗碼，高位在前，低位在后

ID:584814 · 發(fā)表于 2021-4-14 00:11

只能胡猜

ID:319585 · 發(fā)表于 2021-4-14 07:11

modbus算法，樓主詳細了解一下。

ID:247164 · 發(fā)表于 2021-4-14 09:15

采用的是CRC-16/XMODEM校驗方式，F(xiàn)D為幀頭，不參于校驗，數(shù)據(jù)位從88開始，F(xiàn)C為幀尾，前兩位為校驗碼，高位在前，低位在后�？梢杂迷诰€校驗查看，網(wǎng)址為：見下圖

ID:247164 · 發(fā)表于 2021-4-14 09:23

計算參考：【CRC筆記】CRC-16 XMODEM C語言實現(xiàn) https://blog.csdn.net/u012028275/article/details/112067463

ID:905171 · 發(fā)表于 2021-4-14 09:23

應該是crc16 ，反推只能窮舉咯。

ID:619688 · 發(fā)表于 2021-4-14 09:56

夢想未央發(fā)表于 2021-4-13 15:50
采用的是CRC-16/XMODEM校驗方式，F(xiàn)D為幀頭，不參于校驗，數(shù)據(jù)位從88開始，F(xiàn)C為幀尾，前兩位為校驗碼，高 ...

我昨天在網(wǎng)上查到了，也解也來了，是CRC16-CCITT算法，還是謝謝你，分給你了

帳號		自動登錄	找回密碼
密碼			立即注冊

逆向解析單片機串口校驗算法,幀尾FC的前2個字節(jié)是怎么算出來的?

最佳答案

瀏覽過的版塊