CRC24Q 实现
我正在尝试实现 CRC 检查算法，该算法基本上根据输入消息创建一个值。 因此，假设我有一条十六进制消息 3F214365876616AB15387D5D59，并且我想获取该消息的 CRC24Q 值。 我发现执行此操作的算法如下：
typedef unsigned long crc24;
crc24 crc_check(unsigned char *input) {
unsigned char *octets;
crc24 crc = 0xb704ce; // CRC24_INIT;
int i;
int len = strlen(input);
octets = input;
while (len--) {
crc ^= ((*octets++) << 16);
for (i = 0; i < 8; i++) {
crc <<= 1;
if (crc & 0x1000000)
crc ^= CRC24_POLY;
}
}
return crc & 0xFFFFFF;
}
其中*input=3F214365876616AB15387D5D59。 问题是 ((*octets++) << 16) 会将十六进制字符的 ascii 值移动 16 位，而不是字符本身。 因此，我创建了一个将十六进制数字转换为字符的函数。 我知道这个实现看起来很奇怪，如果它是错误的，我也不会感到惊讶。 这是转换函数：
char* convert(unsigned char* message) {
unsigned char* input;
input = message;
int p;
char *xxxx[20];
xxxx[0]="";
for (p = 0; p < length(message) - 1; p = p + 2) {
char* pp[20];
pp[0] = input[0];
char *c[20];
*input++;
c[0]= input[0];
*input++;
strcat(pp,c);
char cc;
char tt[2];
cc = (char ) strtol(pp, &pp, 16);
tt[0]=cc;
strcat(xxxx,tt);
}
return xxxx;
}
SO：
unsigned char *msg_hex="3F214365876616AB15387D5D59";
crc_sum = crc_check(convert((msg_hex)));
printf("CRC-sum: %x\n", crc_sum);
非常感谢您的任何建议。
I am trying to implement the algorithm of a CRC check, which basically created a value, based on an input message.
So, consider I have a hex message 3F214365876616AB15387D5D59, and I want to obtain the CRC24Q value of the message.
The algorithm that I found to do this is the following:
typedef unsigned long crc24;
crc24 crc_check(unsigned char *input) {
unsigned char *octets;
crc24 crc = 0xb704ce; // CRC24_INIT;
int i;
int len = strlen(input);
octets = input;
while (len--) {
crc ^= ((*octets++) << 16);
for (i = 0; i < 8; i++) {
crc <<= 1;
if (crc & 0x1000000)
crc ^= CRC24_POLY;
}
}
return crc & 0xFFFFFF;
}
where *input=3F214365876616AB15387D5D59.
The problem is that ((*octets++) << 16) will shift by 16 bits the ascii value of the hex character and not the character itself.
So, I made a function to convert the hex numbers to characters.
I know the implementation looks weird, and I wouldn't be surprised if it were wrong.
This is the convert function:
char* convert(unsigned char* message) {
unsigned char* input;
input = message;
int p;
char *xxxx[20];
xxxx[0]="";
for (p = 0; p < length(message) - 1; p = p + 2) {
char* pp[20];
pp[0] = input[0];
char *c[20];
*input++;
c[0]= input[0];
*input++;
strcat(pp,c);
char cc;
char tt[2];
cc = (char ) strtol(pp, &pp, 16);
tt[0]=cc;
strcat(xxxx,tt);
}
return xxxx;
}
SO:
unsigned char *msg_hex="3F214365876616AB15387D5D59";
crc_sum = crc_check(convert((msg_hex)));
printf("CRC-sum: %x\n", crc_sum);
Thank you very much for any suggestions.
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if (crc & 0x8000000)不应该是if (crc & 0x1000000)否则你测试的是第 28 位而不是第 25 位的 24 位溢出Shouldn't the
if (crc & 0x8000000)beif (crc & 0x1000000)otherwise you're testing the 28th bit not the 25th for 24-bit overflow