链路层分组汉明码纠错计算原理Hamming Code - data link

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A.以下在接收方接收分组时候产品随即一位错误情况:

Enter the data[max:30]:10101000111
Sender: send data:  10101000111 [Hamming code count:4]
data_with position: __1_010_1000111
datar with Hamming: 001001001000111
Receiver get data:     001001001010111
Receiver: data invalid![1011]
Receiver revise data:001001001000111

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B.以下是在接收方接收分组正确情况:

Enter the data[max:30]:10100101011
Sender: send data:  10100101011 [Hamming code count:4]
data_with position: __1_010_0101011
datar with Hamming: 011001000101011
Receiver get data:     011001000101011
Receiver: data valid!
 

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/*Hamming code* 汉明码通过在分组中添加冗余校验码来实现差错检测和纠正* 实现原理是通过对与某个校验码对应的比特组进行奇偶校验* 其主要的步骤包括:* 1.计算冗余码个数 pow(2,r) >= d + r + 1.  * 注:r代表冗余码数量,d代表分组比特数)* 2.确定冗余码位置. * 注:汉明码校验码对应的位置值转化为二进制后类似于(0001,0010,0100..)*    即2的0次幂,2的1次幂,2的2次幂...* 3.确定冗余码值* 注:冗余码比特值是为了确保该值与同组中的比特值求异或后结果为0*    而组中的成员比特的位置选择遵循以下规则:*    同组比特位对应的二进制与对应冗余码二进制进行求或操作后依然为原值.*    例如: 比特位a对应位置是11(二进制1011) * 		    比特位b对应位置是6(二进制110）*         汉明码h位置1(二进制0001)*    a|h -> 1011 | 0001 = 1011(a位置) 则a位置与汉明码h为同组*    b|h -> 110  | 0001 = 111(非b位置) 则b位置与汉明码不为同组*  4.接收方校验和纠错*    注:接收方通过对冗余码从高位到低位的顺序对分组进行校验,*    如果某组得到结果值为1则说明该组有误,由于不同组存在交叉检验比特位情况,*    这样即能得到错误位置的二进制.*    如:汉明码有4位,则0000代表正确,如0110(高位到低位)则说明第6位比特值有误,将其取反即可* * 注:汉明码只能发现和修正一位错误，对于两位或者两位以上的错误无法正确和发现.*    所以有时候会把大的分组进行拆分，通过对拆分后的组进行对应的汉明码查错和纠错* */#include <math.h>#include <string.h>#include <stdlib.h>#include <stdio.h>#include <stdbool.h>#include <regex.h>#include <assert.h>#include <time.h>typedef int errno_t;//functions prototype//步骤1: 计算校验码个数int nHammingCode(char* data);//步骤2 步骤3:将汉明码插入数据并赋值char* insertHammingCode(int r,char* data);//步骤4:接收方查错纠错bool checkData(char* datar,int r,char** wrongIndex);//检测输入分组有效性bool valid(char* data);//接收方收到无效数据后修纠错void revise(char* datar,char* wrongIndex);//辅助方法.二进制字符串转int
int sbtoi(char* sb);//辅助函数,用于寻找下一个匹配正则的位置.无匹配返回-1
int regNextMatchIndex(regex_t* regex,char* content);int main(void){char* data = calloc(31,sizeof(char));retry:printf("Enter the data[max:30]:");scanf("%s",data);errno_t state = valid(data);if(state == false){printf("data: %s\n",data);goto retry;}int r = nHammingCode(data);printf("Sender: send data:  %s ",data);printf("[Hamming code count:%d]\n",r);char* datar = insertHammingCode(r,data);printf("datar with Hamming: %s\n",datar);//这里上分组随机产生一位错误
#define WRONG_TEST
#undef WRONG_TEST  //这里作为一个开关.注释则产生一位随即错误
#ifdef WRONG_TESTsrand(time(NULL));int wrong = rand() % (strlen(datar) - 1);datar[wrong] = (datar[wrong] == '1' ? '0' : '1');
#endifchar* wrongIndex = NULL;bool check = checkData(datar,r,&wrongIndex);if(!check){printf("Receiver: data invalid![%s]\n",wrongIndex);revise(datar,wrongIndex);}else{printf("Receiver: data valid!\n");}return 0;
}int nHammingCode(char* data){int r = 1;int d = strlen(data);for(;pow(2,r) < d + r + 1;r ++);return r;//返回冗余码个数
}char* insertHammingCode(int r,char* data){//在对应位置插入汉明冗余码并赋值char* datar = calloc(strlen(data) + r + 1,sizeof(char));memcpy(datar,data,strlen(data));int index = 0;int n = 5;//n代表index转化为字符串后的长度.这里直接设置成5位char* pattern = calloc(11 + n + 1,sizeof(char));regex_t regex;errno_t state = 0;regmatch_t matches[3];//这里已知后续正则表达式直接赋值3char* head = NULL;char* tail = NULL;for(int i = 0 ; i < r ; i ++){index = pow(2,i) - 1;//指针偏移量的位置-1memset(pattern,0,11 + n + 1);//清空pattern内存sprintf(pattern,"^(.{%d})(.*)$",index);state = regcomp(&regex,pattern,REG_EXTENDED);if(state){char* errbuf = calloc(50,sizeof(char));regerror(state,&regex,errbuf,50);fprintf(stderr,"Failed to compile Regex:%s\n""Reason: %s\n",pattern,errbuf);free(errbuf);regfree(&regex);exit(EXIT_FAILURE);}state = regexec(&regex,datar,3,matches,0);assert(state == 0);head = calloc(matches[1].rm_eo - matches[1].rm_so + 1,sizeof(char));tail = calloc(matches[2].rm_eo - matches[2].rm_so + 1,sizeof(char));memcpy(head,datar + matches[1].rm_so,matches[1].rm_eo - matches[1].rm_so);memcpy(tail,datar + matches[2].rm_so,matches[2].rm_eo - matches[2].rm_so);memcpy(datar,head,strlen(head));memcpy(datar + strlen(head),"_",1);memcpy(datar + strlen(head) + 1,tail,strlen(tail));free(head);free(tail);regfree(&regex);}printf("data_with position: %s\n",datar);//找出对应的校验码位置并用下横线标注//以下采用一种比较低效的方法对汉明校验码进行赋值int counts[r];memset(counts,0,r * sizeof(int));int indexs[r];memset(indexs,0,r * sizeof(int));for(int i = 0 ; i < r ; i ++){indexs[i] = pow(2,i);}for(int i = 0 ; i < strlen(datar) ; i ++){for(int j = 0 ; j < r ; j ++){if((i + 1) == indexs[j]){//printf("位置:%d不检测.\n",i + 1);continue;}if(((i + 1) | indexs[j]) ==  (i + 1)){//printf("校验码:%d,校验data坐标:%d 值:%c\n",j+1,i+1,datar[i]);if(datar[i] == '1'){counts[j] ++;}}}}for(int i = 0 ; i < r ; i ++){//printf("第%d个检验码,count:%d\n",i + 1,counts[i]);datar[indexs[i] - 1] = (counts[i] % 2 == 0) ? '0' : '1';}//printf("datar: %s\n",datar);return datar;
}bool checkData(char* datar,int r,char** wrongIndex){//步骤4:接收方查错纠错printf("Receiver get data:  %s\n",datar);int indexs[r];int counts[r];*wrongIndex = calloc(r,sizeof(char));memset(counts,0,sizeof(int) * r);for(int i = 0 ; i < r ; i ++){indexs[i] = pow(2,i);}for(int i = 0 ; i < strlen(datar) ; i ++){for(int j = 0 ; j < r ; j ++){if(((i + 1) | indexs[j]) == (i + 1)){//printf("校验码:%d,校验data坐标:%d 值:%c\n",j+1,i+1,datar[i]);if(datar[i] == '1'){counts[j] ++;}}}}bool reval = true;int remainder = 0;for(int i = 0 ; i < r ; i ++){//printf("第%d个检验码,count:%d\n",i + 1,counts[i]);remainder = counts[i] % 2;if(remainder){reval = false;}(*wrongIndex)[r-i - 1] = (remainder == 0 ? '0' : '1');}return reval;
}void revise(char* datar,char* wrongIndex){int index = sbtoi(wrongIndex) - 1;datar[index] = (datar[index] == '1' ? '0' : '1');printf("Receiver revise data:%s\n",datar);
}//辅助方法.二进制字符串转int
int sbtoi(char* sb){regex_t regex;char* pattern = "^[01]+$";errno_t state = regcomp(&regex,pattern,REG_EXTENDED);assert(state == 0);state = regexec(&regex,sb,0,NULL,0);if(state == REG_NOMATCH){fprintf(stderr,"Invalid bit str:%s\n",sb);exit(EXIT_FAILURE);}regfree(&regex);int reval = 0;int len = strlen(sb);pattern = "1";state = regcomp(&regex,pattern,REG_EXTENDED);assert(state == 0);int index = 0;int exponent = 0;while((index = regNextMatchIndex(&regex,sb)) != -1){exponent = len - 1 - index;reval += pow(2,exponent);}regfree(&regex);return reval;
}//辅助函数,用于寻找下一个匹配正则的位置.无匹配返回-1
int regNextMatchIndex(regex_t* regex,char* content){static int position = 0;regmatch_t matches[regex->re_nsub + 1];errno_t state = regexec(regex,content + position,regex->re_nsub + 1,matches,0);if(state == REG_NOMATCH){position = 0;return -1;}int reval = position + matches[0].rm_so;position += matches[0].rm_eo;return reval;
}bool valid(char* data){if(strlen(data) > 30){fprintf(stderr,"MAX: 30 < Len: %zd >>",strlen(data));return false;}regex_t regex;char* pattern = "^[01]+$";errno_t state = regcomp(&regex,pattern,REG_EXTENDED);if(state){char* errbuf = calloc(50,sizeof(char));regerror(state,&regex,errbuf,50);fprintf(stderr,"Failed to compile Regex:%s\n""Reason:%s\n",pattern,errbuf);free(errbuf);regfree(&regex);exit(EXIT_FAILURE);}state = regexec(&regex,data,0,NULL,0);if(state == REG_NOMATCH){fprintf(stderr,"Content Invalid. >>");return false;}return true;
}

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