在C ++中将24位整数(2s补码)转换为32位整数 [英] Converting 24 bit integer (2s complement) to 32 bit integer in C++

问题描述

dataFile.bin是具有6个字节记录的二进制文件.前3个 每条记录的字节包含纬度，而后3个字节包含 经度.每个24位值表示弧度乘以 0X1FFFFF

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

这是我一直在努力的任务.我已经好几年没有做C ++了，所以它花费我的时间比我认为的要长-_-.谷歌搜索后，我看到了对我来说有意义的算法.

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;  
}  

问题是语法问题，我只能按照其他人对此进行编程的方式进行工作.我不明白如何将CHAR数据"存储到INT中.如果数据"是一个数组，那会更有意义吗?由于其接收24个整数信息存储到BYTE中.

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;
}

我非常感谢您的帮助或提示，即使您不了解某个提示或提示也会对我有很大帮助.我只需要和某人交谈.

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.

推荐答案

我不了解如何将CHAR数据"存储到INT中.

I am not understanding how I can store the CHAR "data" into an INT.

由于char是数字类型，因此将它们组合成单个int毫无问题.

Since char is a numeric type, there is no problem combining them into a single int.

由于它接收到存储在BYTE中的24个整数信息

Since its receiving 24 integers of information stored into a BYTE

它是24位而不是字节，因此仅需要组合三个整数值.

It's 24 bits, not bytes, so there are only three integer values that need to be combined.

在不使用条件的情况下产生相同结果的更简单方法如下:

An easier way of producing the same result without using conditionals is as follows:

int interpret24bitAsInt32(byte[] byteArray) {     
    return (  
        (byteArray[0] << 24)
    |   (byteArray[1] << 16)
    |   (byteArray[2] << 8)
    ) >> 8;  
}

这个想法是将作为输入提供的三个字节存储到四字节int的 upper 中的三个字节中，然后将其向下移位一个字节.这样，程序将自动对您的号码进行符号扩展，从而避免了有条件的执行.

The idea is to store the three bytes supplied as an input into the upper three bytes of the four-byte int, and then shift it down by one byte. This way the program would sign-extend your number automatically, avoiding conditional execution.

注意:该代码不可移植，因为它假定32位整数大小.要使其可移植，请使用 <cstdint> 类型:

Note: This code is not portable, because it assumes 32-bit integer size. To make it portable use <cstdint> types:

int32_t interpret24bitAsInt32(int8_t[] byteArray) {
    return (  
        (byteArray[0] << 24)
    |   (byteArray[1] << 16)
    |   (byteArray[2] << 8)
    ) >> 8; 
}

还假定24位数字的最高有效字节存储在byteArray的初始元素中，然后是中间元素，最后是最低有效字节.

It also assumes that the most significant byte of the 24-bit number is stored in the initial element of byteArray, then comes the middle element, and finally the least significant byte.

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