如何在C++中实现十六进制到IEEE75432位浮点数的转换

Question

我正在尝试转换存储为int的十六进制值，并使用IEEE 32位规则将它们转换为浮点数。我特别努力为尾数和指数获得正确的值。十六进制存储在一个十六进制的文件中。我想要有四个重要的数字。下面是我的代码。

float floatizeMe(unsigned int myNumba ) {
    //// myNumba comes in as 32 bits or 8 byte  

    unsigned int  sign = (myNumba & 0x007fffff) >>31;
    unsigned int  exponent = ((myNumba & 0x7f800000) >> 23)- 0x7F;
    unsigned int  mantissa = (myNumba & 0x007fffff) ;
    float  value = 0;
    float mantissa2; 

    cout << endl<< "mantissa is : " << dec << mantissa << endl;

    unsigned    int m1 = mantissa & 0x00400000 >> 23;
    unsigned    int m2 = mantissa & 0x00200000 >> 22;
    unsigned    int m3 = mantissa & 0x00080000 >> 21;
    unsigned    int m4 = mantissa & 0x00040000 >> 20;

    mantissa2 = m1 * (2 ^ -1) + m2*(2 ^ -2) + m3*(2 ^ -3) + m4*(2 ^ -4);

    cout << "\nsign is: " << dec << sign << endl;
    cout << "exponent is : " << dec << exponent << endl;
    cout << "mantissa  2 is : " << dec << mantissa2 << endl;

    // if above this number it is negative 
    if ( sign  == 1)
        sign = -1; 

    // if above this number it is positive 
    else {
        sign = 1;
    }

    value = (-1^sign) * (1+mantissa2) * (2 ^ exponent);
    cout << dec << "Float value is: " << value << "\n\n\n";

    return value;
}




  int main()
{   
    ifstream myfile("input.txt");
    if (myfile.is_open())
    {
        unsigned int a, b,b1; // Hex 
        float c, d, e; // Dec
        int choice; 

        unsigned int ex1 = 0;
        unsigned int ex2 = 1;
        myfile >> std::hex;
        myfile >> a >> b ;
        floatizeMe(a);
myfile.close();
return 0;

}

Answer 1

我猜想你是指^在

mantissa2 = m1 * (2 ^ -1) + m2*(2 ^ -2) + m3*(2 ^ -3) + m4*(2 ^ -4);

意思是“到…的力量”。C或C++中没有这样的运算符。^运算符是逐位异或运算符。

Answer 2

考虑到您的CPU遵循IEEE标准，您也可以使用union。像这样的东西

  union
  {
    int num;
    float fnum;
  } my_union;

然后将整数值存储到my_union.num中，并通过获取my_union.fnum将它们读取为浮点数。

Answer 3

我们需要转换IEEE-754的单精度和双精度数字(使用32位和64位编码)。我们使用C编译器(Vector CANoe/C分析器CAPL脚本)和一组有限的函数，并最终开发了下面的函数(它可以很容易地使用任何on-line C compiler进行测试)：

#include <stdio.h>
#include <math.h>

double ConvertNumberToFloat(unsigned long number, int isDoublePrecision)
{
    int mantissaShift = isDoublePrecision ? 52 : 23;
    unsigned long exponentMask = isDoublePrecision ? 0x7FF0000000000000 : 0x7f800000;
    int bias = isDoublePrecision ? 1023 : 127;
    int signShift = isDoublePrecision ? 63 : 31;

    int sign = (number >> signShift) & 0x01;
    int exponent = ((number & exponentMask) >> mantissaShift) - bias;

    int power = -1;
    double total = 0.0;
    for ( int i = 0; i < mantissaShift; i++ )
    {
        int calc = (number >> (mantissaShift-i-1)) & 0x01;
        total += calc * pow(2.0, power);
        power--;
    }
    double value = (sign ? -1 : 1) * pow(2.0, exponent) * (total + 1.0);

    return value;
}

int main()
{
    // Single Precision 
    unsigned int singleValue = 0x40490FDB; // 3.141592...
    float singlePrecision = (float)ConvertNumberToFloat(singleValue, 0);
    printf("IEEE754 Single (from 32bit 0x%08X): %.7f\n",singleValue,singlePrecision);

    // Double Precision
    unsigned long doubleValue = 0x400921FB54442D18; // 3.141592653589793... 
    double doublePrecision = ConvertNumberToFloat(doubleValue, 1);
    printf("IEEE754 Double (from 64bit 0x%016lX): %.16f\n",doubleValue,doublePrecision);
}