Question

I’m trying (without much success) to write a short c++ function:

double digit(double x, int b, int d)

that returns the d-th digit in base-b expansion of the number x, which can be positive or negative, and can be a fraction. when d is negative, it should return the after-the-decimal-dot digits (its underfined for d=0 so say it returns 0 in that case).
For example:

    const double x = 25.73;
    for (int n = -5; n <= 5; n++)
            cout<<digit(x,10,n)<<' ';

should print:
0
0
0
3
7
0
5
2
0
0
0

The function must use only loops, if, exp, pow, log, floor and ceil. i.e., without sprintf tricks etc.

Thanks!!!

EDIT: For simplicity, assume 2<=b<=10

EDIT: Please also avoid using mod, only pow-exp-log-floor-ceil based solutions

Answer 1

This seems to be the most straightforward implementation, and it seems to work just fine.

int digit( double x, int base, int index ) {
    // shift number (mult by power of base) so desired digit is in one's place
    x = std::abs( x ) * std::pow( base, - index );
    // fmod strips higher digits; floor strips lower digits, leaving result.
    return std::floor( std::fmod( x, base ) );
}

I changed the return type from double to int since it doesn’t make sense to have a fraction in a digit. And it doesn’t return . for 0, because again that’s not a digit. The 0’th place value is the one’s place.

Also this ignores the minus sign; you didn’t define the “base-b expansion” for negative numbers. You could adjust the function to return b’s complement notation or whatever.

By substituting x you can make this into one line, so it will satisfy the requirements for constexpr on platforms where the math functions are constexpr as well.