Converting 24 bit integer (2s complement) to 32 bit integer in C++

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The dataFile.bin is a binary file with 6-byte records. The first 3 bytes of each record contain the latitude and the last 3 bytes contain the longitude. Each 24 bit value represents radians multiplied by 0X1FFFFF

This is a task I've been working on. I havent done C++ in years so its taking me way longer than I thought it would -_-. After googling around I saw this algorthim which made sense to me.

int interpret24bitAsInt32(byte[] byteArray) {     
 int newInt = (  
     ((0xFF & byteArray[0]) << 16) |  
     ((0xFF & byteArray[1]) << 8) |   
     (0xFF & byteArray[2])  
   );  
 if ((newInt & 0x00800000) > 0) {  
   newInt |= 0xFF000000;  
 } else {  
   newInt &= 0x00FFFFFF;  
 }  
return newInt;  
}  

The problem is a syntax issue I am restricting to working by the way the other guy had programmed this. I am not understanding how I can store the CHAR "data" into an INT. Wouldn't it make more sense if "data" was an Array? Since its receiving 24 integers of information stored into a BYTE.

double BinaryFile::from24bitToDouble(char *data) {
    int32_t iValue;

    // ****************************
    // Start code implementation
    // Task: Fill iValue with the 24bit integer located at data.
    // The first byte is the LSB.
    // ****************************
//iValue += 
    // ****************************
    // End code implementation
    // ****************************
    return static_cast<double>(iValue) / FACTOR;
}

bool BinaryFile::readNext(DataRecord &record)
{
    const size_t RECORD_SIZE = 6;
    char buffer[RECORD_SIZE];
    m_ifs.read(buffer,RECORD_SIZE);
    if (m_ifs) {
        record.latitude = toDegrees(from24bitToDouble(&buffer[0]));
        record.longitude = toDegrees(from24bitToDouble(&buffer[3]));
        return true;
    }
    return false;
}

double BinaryFile::toDegrees(double radians) const
{
    static const double PI = 3.1415926535897932384626433832795;
    return radians * 180.0 / PI;
}

I appreciate any help or hints even if you dont understand a clue or hint will help me alot. I just need to talk to someone.

3 Answers
    int32_t upperByte   = ((int32_t) dataRx[0] << 24);
    int32_t middleByte  = ((int32_t) dataRx[1] << 16);
    int32_t lowerByte   = ((int32_t) dataRx[2] << 8);

    int32_t ADCdata32 = (((int32_t) (upperByte | middleByte | lowerByte)) >> 8);     // Right-shift of signed data maintains signed bit
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