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c++ - Swift-来自C++的CRC8计算转换

转载 作者:行者123 更新时间:2023-12-02 09:57:46 24 4
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作为我正在从蓝牙设备提取数据的项目的一部分,制造商为我提供了CRC8函数供使用。我正在用Swift编写应用程序,但是它们提供的功能是在C++中。
以下是他们在C++中提供的功能

unsigned char xTable_CRC8[]={0x00, 0x07, 0x0E, 0x09, 0x1C, 0x1B, 0x12, 0x15, 0x38, 0x3F, 0x36, 0x31, 0x24, 0x23, 0x2A, 0x2D,0x70, 0x77, 0x7E, 0x79, 0x6C, 0x6B, 0x62, 0x65,0x48, 0x4F, 0x46, 0x41, 0x54, 0x53, 0x5A, 0x5D,0xE0, 0xE7, 0xEE, 0xE9, 0xFC, 0xFB, 0xF2, 0xF5,0xD8, 0xDF, 0xD6, 0xD1, 0xC4, 0xC3, 0xCA, 0xCD,0x90, 0x97, 0x9E, 0x99, 0x8C, 0x8B, 0x82, 0x85,0xA8, 0xAF, 0xA6, 0xA1, 0xB4, 0xB3, 0xBA, 0xBD,0xC7, 0xC0, 0xC9, 0xCE, 0xDB, 0xDC, 0xD5, 0xD2,0xFF, 0xF8, 0xF1, 0xF6, 0xE3, 0xE4, 0xED, 0xEA,0xB7, 0xB0, 0xB9, 0xBE, 0xAB, 0xAC, 0xA5, 0xA2,0x8F, 0x88, 0x81, 0x86, 0x93, 0x94, 0x9D, 0x9A,0x27, 0x20, 0x29, 0x2E, 0x3B, 0x3C, 0x35, 0x32,0x1F, 0x18, 0x11, 0x16, 0x03, 0x04, 0x0D, 0x0A,0x57, 0x50, 0x59, 0x5E, 0x4B, 0x4C, 0x45, 0x42,0x6F, 0x68, 0x61, 0x66, 0x73, 0x74, 0x7D, 0x7A,0x89, 0x8E, 0x87, 0x80, 0x95, 0x92, 0x9B, 0x9C,0xB1, 0xB6, 0xBF, 0xB8, 0xAD, 0xAA, 0xA3, 0xA4,0xF9, 0xFE, 0xF7, 0xF0, 0xE5, 0xE2, 0xEB, 0xEC,0xC1, 0xC6, 0xCF, 0xC8, 0xDD, 0xDA, 0xD3, 0xD4,0x69, 0x6E, 0x67, 0x60, 0x75, 0x72, 0x7B, 0x7C,0x51, 0x56, 0x5F, 0x58, 0x4D, 0x4A, 0x43, 0x44,0x19, 0x1E, 0x17, 0x10, 0x05, 0x02, 0x0B, 0x0C,0x21, 0x26, 0x2F, 0x28, 0x3D, 0x3A, 0x33, 0x34,0x4E, 0x49, 0x40, 0x47, 0x52, 0x55, 0x5C, 0x5B,0x76, 0x71, 0x78, 0x7F, 0x6A, 0x6D, 0x64, 0x63,0x3E, 0x39, 0x30, 0x37, 0x22, 0x25, 0x2C, 0x2B,0x06, 0x01, 0x08, 0x0F, 0x1A, 0x1D, 0x14, 0x13,0xAE, 0xA9, 0xA0, 0xA7, 0xB2, 0xB5, 0xBC, 0xBB,0x96, 0x91, 0x98, 0x9F, 0x8A, 0x8D, 0x84, 0x83,0xDE, 0xD9, 0xD0, 0xD7, 0xC2, 0xC5, 0xCC, 0xCB,0xE6, 0xE1, 0xE8, 0xEF, 0xFA, 0xFD, 0xF4, 0xF3 };

uint8_t CRC8(char *RP_ByteData, unsigned int Buffer_Size) {

uint8_t x,R_CRC_Data;

R_CRC_Data=0;

for(unsigned int i = 0; i < Buffer_Size; i++) {
x = R_CRC_Data ^ (*RP_ByteData);
R_CRC_Data = xTable_CRC8[x];
RP_ByteData++;
}

return R_CRC_Data;
}
这是到目前为止我尝试进行的转换:
  func calCR8(buf : [UInt8]) -> UInt8 {
let initialValue : UInt8 = UInt8.min;
let Table_CRC8 : [UInt8] = [
0x00, 0x07, 0x0E, 0x09, 0x1C, 0x1B, 0x12, 0x15,0x38, 0x3F, 0x36, 0x31, 0x24, 0x23, 0x2A, 0x2D,
0x70, 0x77, 0x7E, 0x79, 0x6C, 0x6B, 0x62, 0x65,0x48, 0x4F, 0x46, 0x41, 0x54, 0x53, 0x5A, 0x5D,
0xE0, 0xE7, 0xEE, 0xE9, 0xFC, 0xFB, 0xF2, 0xF5, 0xD8, 0xDF, 0xD6, 0xD1, 0xC4, 0xC3, 0xCA, 0xCD,
0x90, 0x97, 0x9E, 0x99, 0x8C, 0x8B, 0x82, 0x85,0xA8, 0xAF, 0xA6, 0xA1, 0xB4, 0xB3, 0xBA, 0xBD,
0xC7, 0xC0, 0xC9, 0xCE, 0xDB, 0xDC, 0xD5, 0xD2, 0xFF, 0xF8, 0xF1, 0xF6, 0xE3, 0xE4, 0xED, 0xEA,
0xB7, 0xB0, 0xB9, 0xBE, 0xAB, 0xAC, 0xA5, 0xA2,0x8F, 0x88, 0x81, 0x86, 0x93, 0x94, 0x9D, 0x9A,
0x27, 0x20, 0x29, 0x2E, 0x3B, 0x3C, 0x35, 0x32,0x1F, 0x18, 0x11, 0x16, 0x03, 0x04, 0x0D, 0x0A,
0x57, 0x50, 0x59, 0x5E, 0x4B, 0x4C, 0x45, 0x42,0x6F, 0x68, 0x61, 0x66, 0x73, 0x74, 0x7D, 0x7A,
0x89, 0x8E, 0x87, 0x80, 0x95, 0x92, 0x9B, 0x9C,0xB1, 0xB6, 0xBF, 0xB8, 0xAD, 0xAA, 0xA3, 0xA4,
0xF9, 0xFE, 0xF7, 0xF0, 0xE5, 0xE2, 0xEB, 0xEC, 0xC1, 0xC6, 0xCF, 0xC8, 0xDD, 0xDA, 0xD3, 0xD4,
0x69, 0x6E, 0x67, 0x60, 0x75, 0x72, 0x7B, 0x7C,0x51, 0x56, 0x5F, 0x58, 0x4D, 0x4A, 0x43, 0x44,
0x19, 0x1E, 0x17, 0x10, 0x05, 0x02, 0x0B, 0x0C,0x21, 0x26, 0x2F, 0x28, 0x3D, 0x3A, 0x33, 0x34,
0x4E, 0x49, 0x40, 0x47, 0x52, 0x55, 0x5C, 0x5B,0x76, 0x71, 0x78, 0x7F, 0x6A, 0x6D, 0x64, 0x63,
0x3E, 0x39, 0x30, 0x37, 0x22, 0x25, 0x2C, 0x2B,0x06, 0x01, 0x08, 0x0F, 0x1A, 0x1D, 0x14, 0x13,
0xAE, 0xA9, 0xA0, 0xA7, 0xB2, 0xB5, 0xBC, 0xBB,0x96, 0x91, 0x98, 0x9F, 0x8A, 0x8D, 0x84, 0x83,
0xDE, 0xD9, 0xD0, 0xD7, 0xC2, 0xC5, 0xCC, 0xCB, 0xE6, 0xE1, 0xE8, 0xEF, 0xFA, 0xFD, 0xF4, 0xF3 ];


var crc = initialValue;
let x: UInt8

for i in 0 ..< buf.count{
x = crc ^ buf
crc = Table_CRC8[x]
}

return crc;
}
}
我正在处理两个类型错误:无法将类型[UInt8]的值转换为期望的参数类型UInt8,并且无法将类型UInt8的值转换为期望的参数类型Int。我在这里感到很失落和不舒服。
基本上,将以如下方式调用此函数:
var buf : [UInt8] = [0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00];
buf[0] = 0xAA;
buf[1] = 0x14;
buf[2] = ~0x14
buf[buf.count - 1] = self.calcCR8(buf)
基于此字节数组在本示例中使用参数,我期望返回0xC6。

最佳答案

您有一些小错误。首先,您尝试对buf执行xor计算-这是整个数组;您需要访问一个元素buf[i]
第二个问题是您无法使用UInt8索引序列,因此需要访问等效的Int才能访问Table_CRC8数组。
最后,您将x声明为let循环之外的for(常数),因此无法进行更改。您可以将x声明移到for循环内以解决此问题。
进行这些小的更改将使您的代码可以编译:

for i in 0 ..< buf.count {
let x = crc ^ buf[i]
crc = Table_CRC8[Int(x)]
}
您可以通过更改CRC表的名称以使其与Swift习惯用法保持一致,并使用迭代循环而不是 for in <range>循环来使其更加“Swifty”:
func calcCRC8(_ buf : [UInt8]) -> UInt8 {
let tableCRC8 : [UInt8] = [
0x00, 0x07, 0x0E, 0x09, 0x1C, 0x1B, 0x12, 0x15,0x38, 0x3F, 0x36, 0x31, 0x24, 0x23, 0x2A, 0x2D,
0x70, 0x77, 0x7E, 0x79, 0x6C, 0x6B, 0x62, 0x65,0x48, 0x4F, 0x46, 0x41, 0x54, 0x53, 0x5A, 0x5D,
0xE0, 0xE7, 0xEE, 0xE9, 0xFC, 0xFB, 0xF2, 0xF5, 0xD8, 0xDF, 0xD6, 0xD1, 0xC4, 0xC3, 0xCA, 0xCD,
0x90, 0x97, 0x9E, 0x99, 0x8C, 0x8B, 0x82, 0x85,0xA8, 0xAF, 0xA6, 0xA1, 0xB4, 0xB3, 0xBA, 0xBD,
0xC7, 0xC0, 0xC9, 0xCE, 0xDB, 0xDC, 0xD5, 0xD2, 0xFF, 0xF8, 0xF1, 0xF6, 0xE3, 0xE4, 0xED, 0xEA,
0xB7, 0xB0, 0xB9, 0xBE, 0xAB, 0xAC, 0xA5, 0xA2,0x8F, 0x88, 0x81, 0x86, 0x93, 0x94, 0x9D, 0x9A,
0x27, 0x20, 0x29, 0x2E, 0x3B, 0x3C, 0x35, 0x32,0x1F, 0x18, 0x11, 0x16, 0x03, 0x04, 0x0D, 0x0A,
0x57, 0x50, 0x59, 0x5E, 0x4B, 0x4C, 0x45, 0x42,0x6F, 0x68, 0x61, 0x66, 0x73, 0x74, 0x7D, 0x7A,
0x89, 0x8E, 0x87, 0x80, 0x95, 0x92, 0x9B, 0x9C,0xB1, 0xB6, 0xBF, 0xB8, 0xAD, 0xAA, 0xA3, 0xA4,
0xF9, 0xFE, 0xF7, 0xF0, 0xE5, 0xE2, 0xEB, 0xEC, 0xC1, 0xC6, 0xCF, 0xC8, 0xDD, 0xDA, 0xD3, 0xD4,
0x69, 0x6E, 0x67, 0x60, 0x75, 0x72, 0x7B, 0x7C,0x51, 0x56, 0x5F, 0x58, 0x4D, 0x4A, 0x43, 0x44,
0x19, 0x1E, 0x17, 0x10, 0x05, 0x02, 0x0B, 0x0C,0x21, 0x26, 0x2F, 0x28, 0x3D, 0x3A, 0x33, 0x34,
0x4E, 0x49, 0x40, 0x47, 0x52, 0x55, 0x5C, 0x5B,0x76, 0x71, 0x78, 0x7F, 0x6A, 0x6D, 0x64, 0x63,
0x3E, 0x39, 0x30, 0x37, 0x22, 0x25, 0x2C, 0x2B,0x06, 0x01, 0x08, 0x0F, 0x1A, 0x1D, 0x14, 0x13,
0xAE, 0xA9, 0xA0, 0xA7, 0xB2, 0xB5, 0xBC, 0xBB,0x96, 0x91, 0x98, 0x9F, 0x8A, 0x8D, 0x84, 0x83,
0xDE, 0xD9, 0xD0, 0xD7, 0xC2, 0xC5, 0xCC, 0xCB, 0xE6, 0xE1, 0xE8, 0xEF, 0xFA, 0xFD, 0xF4, 0xF3 ];


var crc = UInt8.min

for byte in buf {
let x = crc ^ byte
crc = tableCRC8[Int(x)]
}

return crc
}

关于c++ - Swift-来自C++的CRC8计算转换,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64379287/

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