java - 用java计算LRC

Question

我想计算以下消息的 LRC:

T = 0101 0100P = 0101 00001 = 0011 00012 = 0011 0010Starting with 0x00 as the initial byte.0 XOR ‘T’:         0000 00000101 0100Result, LRC =    0101 0100LRC XOR ‘P’:    0101 0100                        0101 0000Result, LRC =    0000 0100LRC XOR ‘1’:    0000 0100                        0011 0001Result, LRC =    0011 0101LRC XOR ‘2’:    0011 0101                        0011 0010Result, LRC =    0000 0111

The code i have tried so far is shown below:

public class TexttoBinary {

private static final int maxBytes = 1;

public static int convert(char number) {

   // BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    System.out.print("Type the number to parse: ");
    //int number = Integer.parseInt(in.readLine());
    int Bit;
    //char number = '1';
    //UInt8 i;
    String result = "";
    for (int i = maxBytes * 7; i >=0; i--) {
        Bit = 1 << i;
        if (number >= Bit) {
            result += 1;
            //System.out.println(result);
            number -= Bit;
        } else {
            result += 0;
        }
    }
    System.out.println(result);
   return Integer.parseInt(result);
}
public static void main(String args[]){
    String msg ="TP12";
    char[] toCharArray = msg.toCharArray();
    char lrc=0;
    for(int i=0;i<toCharArray.length;i++)
    lrc ^=convert(toCharArray[i]);
    System.out.println((byte)lrc);
}

}

我得到的输出是不同的。我现在该怎么办？

Answer 1

假设您正在谈论纵向冗余检查；如有疑问，check you algorithm against published examples .

Here's one in Java ，这看起来和你的很不一样。也许您尝试使用 DeMorgan 定理来优化 LRC，但很有可能它在此过程中发现了错误。