串口常用校验算法（累加和异或和）

Posted 2021-01-26 liuguan

1、异或和

　　异或和就是对数据逐一异或计算（异或结果与下一个数据异或）

　　下面是校验实例

local rdbuf = "FFFF28000000000000600001000D0000000000000000000000000000000000000000000000000000000096"        
local fast = string.sub(rdbuf, 5, string.len(rdbuf)-2)
local last = string.sub(rdbuf, string.len(rdbuf)-1, string.len(rdbuf))
local num = bit.bxor(tonumber(string.sub(fast, 1, 2), 16), tonumber(string.sub(fast, 3, 4), 16)) --异或
local length = string.len(fast) - 1
for i = 5, length, 2 do
      num = bit.bxor(num, tonumber(string.sub(fast, i, i+1), 16))   
end 
if num == tonumber(last, 16) then
校验成功
end

2、累加和

　　累加和也是逐一累加，但是计算不进位，所以最终结果其实是累加结果求256的余数

　　下面是计算实例

local cmd = 10
cmd = string.format("%02X",cmd) -- 02X 是为了保留两位，空位置0   例如（A是0A）
local fast = "18000000000000600001000200".. cmd .."00000000000000010006"
local length = #fast - 1  
local num = 0
for i = 1, length, 2 do
    num = num + tonumber(string.sub(fast, i, i+1), 16)  
end 
num = num % 256
num = string.format("%02X",num)
local instruction = "FFFF" .. fast .. num

print(instruction)