Convert double to integer mantissa and exponents

Question

I am trying extract the mantissa and exponent part from the double. For the test data '0.15625', expected mantissa and exponent are '5' and '-5' respectively (5*2^-5).

double value = 0.15625; 
double mantissa  = frexp(value, &exp);

Result: mantissa = 0.625 and exp = -2.

Here the mantissa returned is a fraction. For my use case (ASN.1 encoding), mantissa should be integer. I understand by right-shifting the mantissa and adjusting the exponent, I can convert binary fraction to the integer. In the eg, 0.625 base 10 is 0.101 base 2, so 3 bytes to be shifted to get the integer. But I am finding it difficult to find a generic algorithm.

So my question is, how do I calculate the number bits to be shifted to convert a decimal fraction to a binary integer?

Answer 1

#include <cmath>        //  For frexp.
#include <iomanip>      //  For fixed and setprecision.
#include <iostream>     //  For cout.
#include <limits>       //  For properties of floating-point format.


int main(void)
{
    double value = 0.15625;

    //  Separate value into significand in [.5, 1) and exponent.
    int exponent;
    double significand = std::frexp(value, &exponent);

    //  Scale significand by number of digits in it, to produce an integer.
    significand = scalb(significand, std::numeric_limits<double>::digits);

    //  Adjust exponent to compensate for scaling.
    exponent -= std::numeric_limits<double>::digits;

    //  Set stream to print significand in full.
    std::cout << std::fixed << std::setprecision(0);

    //  Output triple with significand, base, and exponent.
    std::cout << "(" << significand << ", "
        << std::numeric_limits<double>::radix << ", " << exponent << ")\n";
}

Sample output:

(5629499534213120, 2, -55)

(If the value is zero, you might wish to force the exponent to zero, for aesthetic reasons. Mathematically, any exponent would be correct.)