三字节汉明码校验程序

秒速五厘米 2022-11-14 11:57 302阅读 0赞

三个字节的汉明码校验程序，linux下编译，存储格式与汉明码有一定区别，前三字节存储数据，第四字节存储校验码。

测试程序，在中间随机改变数据一位，模拟传输位翻转。

#include<stdlib.h>
    #include<stdio.h>
    #include<string.h>
    #include<unistd.h>
    
    #include <time.h>
    
    #define BIT0(x) (x&0x01)
    #define BIT1(x) ((x&0x02)>>1)
    #define BIT2(x) ((x&0x04)>>2)
    #define BIT3(x) ((x&0x08)>>3)
    #define BIT4(x) ((x&0x10)>>4)
    #define BIT5(x) ((x&0x20)>>5)
    #define BIT6(x) ((x&0x40)>>6)
    #define BIT7(x) ((x&0x80)>>7)
    // |-----------------------|-----------------------|-----------------------|-----------------------|
    // |H1|H2|03|H3|05|06|07|H4|09|10|11|12|13|14|15|H5|17|18|19|20|21|22|23|24|25|26|27|28|29|30|31|32|
    //       |A0    A1 A2 A3    A4 A5 A6 A7|B0 B1 B2    B3 B4 B5 B6 B7|C0 C1 C2 C3 C4 C5 C6 C7|           
    // H1 = 03+05+07+09+11+13+15+17+19+21+23+25+27+29                                                       
    //      A0+A1+A3+A4+A6+B0+B2+B3+B5+B7+C1+C3+C5+C7                                                     
    // H2 = 03+06+07+10+11+14+15+18+19+22+23+26+27                                                            
    //      A0+A2+A3+A5+A6+B1+B2+B4+B5+C0+C1+C4+C5                                                       
    // H3 = 05+06+07+12+13+14+15+20+21+22+23+28+29                                                        
    //      A1+A2+A3+A7+B0+B1+B2+B6+B7+C0+C1+C6+C7                                                       
    // H4 = 09+10+11+12+13+14+15+24+25+26+27+28+29                                                          
    //      A4+A5+A6+A7+B0+B1+B2+C2+C3+C4+C5+C6+C7                                                      
    // H5 = 17+18+19+20+21+22+23+24+25+26+27+28+29                                                        
    //      B3+B4+B5+B6+B7+C0+C1+C2+C3+C4+C5+C6+C7                                                       
    // |A0-------------------A7|B0-------------------B7|C0-------------------C7|H1|H2|H3|H4|H5|XX|XX|XX|
    
    unsigned char *code_han(unsigned char a, unsigned char b, unsigned char c)
    {
        unsigned char *out = (unsigned char *)malloc(4);
        unsigned char h1 = BIT0(a) + BIT1(a) + BIT3(a) + BIT4(a) + BIT6(a) + 
                           BIT0(b) + BIT2(b) + BIT3(b) + BIT5(b) + BIT7(b) +
                           BIT1(c) + BIT3(c) + BIT5(c) + BIT7(c);
    
        unsigned char h2 = BIT0(a) + BIT2(a) + BIT3(a) + BIT5(a) + BIT6(a) + 
                           BIT1(b) + BIT2(b) + BIT4(b) + BIT5(b) +
                           BIT0(c) + BIT1(c) + BIT4(c) + BIT5(c);
    
        unsigned char h3 = BIT1(a) + BIT2(a) + BIT3(a) + BIT7(a) +
                           BIT0(b) + BIT1(b) + BIT2(b) + BIT6(b) + BIT7(b) +
                           BIT0(c) + BIT1(c) + BIT6(c) + BIT7(c);
    
        unsigned char h4 = BIT4(a) + BIT5(a) + BIT6(a) + BIT7(a) + 
                           BIT0(b) + BIT1(b) + BIT2(b) +
                           BIT2(c) + BIT3(c) + BIT4(c) + BIT5(c) + BIT6(c) + BIT7(c);
    
        unsigned char h5 = BIT3(b) + BIT4(b) + BIT5(b) + BIT6(b) + BIT7(b) +
                           BIT0(c) + BIT1(c) + BIT2(c) + BIT3(c) + BIT4(c) + BIT5(c) + BIT6(c) + BIT7(c);
    
        out[0] = a;
        out[1] = b;
        out[2] = c;
        out[3] = (BIT0(h1)<<7) | (BIT0(h2)<<6) | (BIT0(h3)<<5) | (BIT0(h4)<<4) | (BIT0(h5)<<3);
    
        return out;
    }
    
    #define HA 0xFF
    #define HB 0xFF
    #define HC 0xFF
    #define H1a 0x5B
    #define H1b 0xAD
    #define H1c 0xAA
    #define H2a 0x6D
    #define H2b 0x36
    #define H2c 0x33
    #define H3a 0x8E
    #define H3b 0xC7
    #define H3c 0xC3
    #define H4a 0xF0
    #define H4b 0x07
    #define H4c 0xFC
    #define H5a 0x00
    #define H5b 0xF8
    #define H5c 0xFF
    #define H1 0x80
    #define H2 0x40
    #define H3 0x20
    #define H4 0x10
    #define H5 0x08
    unsigned char *code_ming(unsigned char *hm)
    {
        unsigned char *out = (unsigned char *)malloc(3);
        unsigned char *han = code_han(hm[0], hm[1], hm[2]);
        unsigned char xy = han[3]^hm[3];
        unsigned char corr = 0, fail = 0xFF;
        if (xy&H1) fail &= H1a;
        else corr |= H1a;
        if (xy&H2) fail &= H2a;
        else corr |= H2a;
        if (xy&H3) fail &= H3a;
        else corr |= H3a;
        if (xy&H4) fail &= H4a;
        else corr |= H4a;
        if (xy&H5) fail &= H5a;
        else corr |= H5a;
        out[0] = (fail&(~corr))^hm[0];
        corr = 0; fail = 0xFF;
        if (xy&H1) fail &= H1b;
        else corr |= H1b;
        if (xy&H2) fail &= H2b;
        else corr |= H2b;
        if (xy&H3) fail &= H3b;
        else corr |= H3b;
        if (xy&H4) fail &= H4b;
        else corr |= H4b;
        if (xy&H5) fail &= H5b;
        else corr |= H5b;
        out[1] = (fail&(~corr))^hm[1];
        corr = 0; fail = 0xFF;
        if (xy&H1) fail &= H1c;
        else corr |= H1c;
        if (xy&H2) fail &= H2c;
        else corr |= H2c;
        if (xy&H3) fail &= H3c;
        else corr |= H3c;
        if (xy&H4) fail &= H4c;
        else corr |= H4c;
        if (xy&H5) fail &= H5c;
        else corr |= H5c;
        out[2] = (fail&(~corr))^hm[2];
    
        return out;
    }
    
    #define random_1(a,b) ((rand()%(b-a))+a)
    void break_one(unsigned char *hm)
    {
        srand((int)time(NULL));
        int a = random_1(0,3);
        int b = random_1(0,8);
        printf("Flip data%d bit%d\n", a, b);
        hm[a] = hm[a]^(1<<b);
    }
    
    int main()
    {
        printf("|data0|data1|data2|han  |\n");
        unsigned char *ret = code_han(0x11, 0x22, 0xd3);
        printf("  %02X    %02X    %02X    %02X\n", ret[0], ret[1], ret[2], ret[3]);
        break_one(ret);
        printf("  %02X    %02X    %02X    %02X\n", ret[0], ret[1], ret[2], ret[3]);
        unsigned char *rret = code_ming(ret);
        printf("After Hamming code is corrected.\n");
        printf("  %02X    %02X    %02X\n", rret[0], rret[1], rret[2]);
        return 0;
    }

运行结果

汉明码还是很局限的，多位翻转就不行了，校验位翻转也不行，所以这个程序就用来参考学习吧。