我想为CRC编写一个程序,不同的标准(位处理)卡在init参数上。当init = 0x00正确工作时,不仅对CRC-8 .但是,一旦您更改了init,值就不会正确地给出。有什么问题吗?Init只需要更改寄存器的初始值?
CRC-8 / init = 0x00,poly= 0x07 -工作良好
CRC-8 CDMA / init = 0xFF,poly= 0x9b - aldeady
“儿童权利公约”第八届会议:
int CRC8() {
dynamic_bitset<> regix = MyCRC::GetRegixAsBits(0x00, 8); // init = 0x00
dynamic_bitset<> mess = MyCRC::GetIntAsBitset(0x41, 8); // mess = 0x41
dynamic_bitset<> poly = MyCRC::GetPolyAsBitset(8, 0x07); // poly = 0x07
cout << regix << endl; // 0000 0000 == 0x00
cout << mess << endl; // 0100 0001 0000 0000 == 0x41 + 8 нулей
cout << poly << endl; // 0000 0111 == 0x07
while (mess.size() > 0) {
if (regix[7] == 0) {
regix = regix << 1;
regix[0] = mess[mess.size() - 1];
}
else {
regix = regix << 1;
regix[0] = mess[mess.size() - 1];
regix = regix ^ poly;
}
mess.pop_back();
}
cout << hex << regix.to_ulong() << endl; // 1100 0000 = 0xC0 | 0xC0 (crccalc.com) OK
return regix.to_ulong();
}CRC-8 CDMA:
int CRC8_CDMA() {
dynamic_bitset<> regix = MyCRC::GetRegixAsBits(0xFF, 8); // init = 0xFF
dynamic_bitset<> mess = MyCRC::GetIntAsBitset(0x41, 8); // mess = 0x41
dynamic_bitset<> poly = MyCRC::GetPolyAsBitset(8, 0x9b); // poly = 0x9b
cout << regix << endl; // 1111 1111 == 0xFF
cout << mess << endl; // 0100 0001 0000 0000 == 0x41 + 8 нулей
cout << poly << endl; // 1001 1011 == 0x9b
while (mess.size() > 0) {
if (regix[7] == 0) {
regix = regix << 1;
regix[0] = mess[mess.size() - 1];
}
else {
regix = regix << 1;
regix[0] = mess[mess.size() - 1];
regix = regix ^ poly;
}
mess.pop_back();
}
cout << hex << regix.to_ulong() << endl; // 1110 0010 = 0xE2 | 0x28 (crccalc.com) FALSE
return regix.to_ulong();
}发布于 2017-05-06 21:46:45
错误操作的顺序。对左移位CRC的操作顺序应该是消息的CRC ^= MSB的MSB (最重要位),如果结果为1,则CRC <<= 1,CRC ^= poly,否则CRC <<= 1。然后在消息的MSB旁边重复该过程,等等。
它也没有显示位的排序对于Get...Bits.。
对于第一种情况,我得到0xC0,对于第二种情况,我得到0x28。
代码可以通过xor‘在8位的时间进行简化:
typedef unsigned char BYTE;
BYTE gencrc1(BYTE *bfr, size_t len)
{
size_t i;
BYTE crc = 0x00;
while(len--){
crc ^= *bfr++;
for(i = 0; i < 8; i++){
if(crc & 0x80){
crc <<= 1;
crc ^= 0x07;
} else {
crc <<= 1;
}
}
}
return(crc);
}
BYTE gencrc2(BYTE *bfr, size_t len)
{
size_t i;
BYTE crc = 0xff;
while(len--){
crc ^= *bfr++;
for(i = 0; i < 8; i++){
if(crc & 0x80){
crc <<= 1;
crc ^= 0x9b;
} else {
crc <<= 1;
}
}
}
return(crc);
}gencrc2示例使用“长手除法”,CRCpoly= 0x19b = 110011011,message = 0x41,附加8个零位(用于剩余)。
11011000
------------------
110011011 | 0100000100000000 0x41 with 8 zero bits
11111111 crc init value is 0xff
--------
101111100
110011011
---------
111001110
110011011
---------
010101010
000000000
---------
101010100
110011011
---------
110011110
110011011
---------
000001010
000000000
---------
000010100
000000000
---------
000101000
000000000
---------
00101000 0x28 is remainderhttps://stackoverflow.com/questions/43823923
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