Question

As part of a numerical library test I need to choose base 10 decimal numbers that can be represented exactly in base 2. How do you detect in C++ if a base 10 decimal number can be represented exactly in base 2?

My first guess is as follows:

bool canBeRepresentedInBase2(const double &pNumberInBase10) {     //check if a number in base 10 can be represented exactly in base 2     //reference: http://en.wikipedia.org/wiki/Binary_numeral_system     bool funcResult = false;      int nbOfDoublings = 16*3;     double doubledNumber = pNumberInBase10;     for (int i = 0; i < nbOfDoublings ; i++)     {         doubledNumber = 2*doubledNumber;         double intPart;         double fracPart = modf(doubledNumber/2, &intPart);         if (fracPart == 0) //number can be represented exactly in base 2         {             funcResult = true;             break;         }     }     return funcResult; } 

I tested this function with the following values: -1.0/4.0, 0.0, 0.1, 0.2, 0.205, 1.0/3.0, 7.0/8.0, 1.0, 256.0/255.0, 1.02, 99.005. It returns true for -1.0/4.0, 0.0, 7.0/8.0, 1.0, 99.005 which is correct.

Any better ideas?

Answer 1

I think what you are looking for is a number which has a fractional portion which is the sum of a sequence of negative powers of 2 (aka: 1 over a power of 2). I believe this should always be able to be represented exactly in IEEE floats/doubles.

For example:

0.375 = (1/4 + 1/8) which should have an exact representation.

If you want to generate these. You could try do something like this:

#include <iostream> #include <cstdlib>  int main() {     srand(time(0));     double value = 0.0;     for(int i = 1; i < 256; i *= 2) {         // doesn't matter, some random probability of including this         // fraction in our sequence..         if((rand() % 3) == 0) {             value += (1.0 / static_cast<double>(i));                 }     }     std::cout << value << std::endl; } 

EDIT: I believe your function has a broken interface. It would be better if you had this:

bool canBeRepresentedExactly(int numerator, int denominator); 

because not all fractions have exact representations, but the moment you shove it into a double, you’ve chosen a representation in binary… defeating the purpose of the test.