C# 如何计算纵向冗余校验 (LRC)？
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How can I calculate Longitudinal Redundancy Check (LRC)?
提问by Alexx Perez
I've tried the example from wikipedia: http://en.wikipedia.org/wiki/Longitudinal_redundancy_check
我已经尝试过维基百科的例子：http: //en.wikipedia.org/wiki/Longitudinal_redundancy_check
This is the code for lrc (C#):
这是 lrc (C#) 的代码：
/// <summary>
/// Longitudinal Redundancy Check (LRC) calculator for a byte array.
/// ex) DATA (hex 6 bytes): 02 30 30 31 23 03
/// LRC (hex 1 byte ): EC
/// </summary>
public static byte calculateLRC(byte[] bytes)
{
byte LRC = 0x00;
for (int i = 0; i < bytes.Length; i++)
{
LRC = (LRC + bytes[i]) & 0xFF;
}
return ((LRC ^ 0xFF) + 1) & 0xFF;
}
It said the result is "EC" but I get "71", what I'm doing wrong?
它说结果是“EC”但我得到“71”，我做错了什么？
Thanks.
谢谢。
采纳答案by harold
Here's a cleaned up version that doesn't do all those useless operations (instead of discarding the high bits every time, they're discarded all at once in the end), and it gives the result you observed. This is the version that uses addition, but that has a negation at the end - might as well subtract and skip the negation. That's a valid transformation even in the case of overflow.
这是一个清理过的版本，它不会执行所有这些无用的操作（而不是每次都丢弃高位，它们最终会一次性全部丢弃），它给出了您观察到的结果。这是使用加法的版本，但最后有一个否定 - 不妨减去并跳过否定。即使在溢出的情况下，这也是一个有效的转换。
public static byte calculateLRC(byte[] bytes)
{
int LRC = 0;
for (int i = 0; i < bytes.Length; i++)
{
LRC -= bytes[i];
}
return (byte)LRC;
}
Here's the alternative LRC (a simple xor of bytes)
这是替代的 LRC（一个简单的字节异或）
public static byte calculateLRC(byte[] bytes)
{
byte LRC = 0;
for (int i = 0; i < bytes.Length; i++)
{
LRC ^= bytes[i];
}
return LRC;
}
And Wikipedia is simply wrong in this case, both in the code (doesn't compile) and in the expected result.
在这种情况下，维基百科在代码（未编译）和预期结果中都是错误的。
回答by BarToK
The corrected Wikipedia version is as follows:
修正后的维基百科版本如下：
private byte calculateLRC(byte[] b)
{
byte lrc = 0x00;
for (int i = 0; i < b.Length; i++)
{
lrc = (byte)((lrc + b[i]) & 0xFF);
}
lrc = (byte)(((lrc ^ 0xff) + 2) & 0xFF);
return lrc;
}
回答by Charlie
I realize that this question pretty old, but I had trouble figuring out how to do this. It's working now, so I figured I should paste the code. In my case, the checksum needs to return as an ASCII string.
我意识到这个问题已经很老了，但我很难弄清楚如何做到这一点。它现在可以工作了，所以我想我应该粘贴代码。就我而言，校验和需要作为 ASCII 字符串返回。
public function getLrc($string)
{
$LRC = 0;
// Get hex checksum.
foreach (str_split($string, 1) as $char) {
$LRC ^= ord($char);
}
$hex = dechex($LRC);
// convert hex to string
$str = '';
for($i=0;$i<strlen($hex);$i+=2) $str .= chr(hexdec(substr($hex,$i,2)));
return $str;
}
回答by Omidoo
Guess this one looks cooler ;)
猜猜这个看起来更酷;)
public static byte calculateLRC(byte[] bytes)
{
return bytes.Aggregate<byte, byte>(0, (x, y) => (byte) (x^ y));
}
回答by amp
If someone wants to get the LRC char from a string:
如果有人想从字符串中获取 LRC 字符：
public static char CalculateLRC(string toEncode)
{
byte[] bytes = Encoding.ASCII.GetBytes(toEncode);
byte LRC = 0;
for (int i = 0; i < bytes.Length; i++)
{
LRC ^= bytes[i];
}
return Convert.ToChar(LRC);
}
回答by Hasala Senevirathne
I created this for Arduino to understand the algorithm (of course it's not written in the most efficient way)
我为 Arduino 创建了这个来理解算法（当然它不是以最有效的方式编写的）
String calculateModbusAsciiLRC(String input)
{
//Refer this document http://www.simplymodbus.ca/ASCII.htm
if((input.length()%2)!=0) { return "ERROR COMMAND SHOULD HAVE EVEN NUMBER OF CHARACTERS"; }
// Make sure to omit the semicolon in input string and input String has even number of characters
byte byteArray[input.length()+1];
input.getBytes(byteArray, sizeof(byteArray));
byte LRC = 0;
for (int i = 0; i <sizeof(byteArray)/2; i++)
{
// Gettting the sum of all registers
uint x=0;
if(47<byteArray[i*2] && byteArray[i*2] <58) {x=byteArray[i*2] -48;}
else { x=byteArray[i*2] -55; }
uint y=0;
if(47<byteArray[i*2+1] && byteArray[i*2+1] <58) {y=byteArray[i*2+1] -48;}
else { y=byteArray[i*2+1] -55; }
LRC += x*16 + y;
}
LRC = ~LRC + 1; // Getting twos Complement
String checkSum = String(LRC, HEX);
checkSum.toUpperCase(); // Converting to upper case eg: bc to BC - Optional somedevices are case insensitve
return checkSum;
}
