Boost cpp_dec_float supports modulus operation

Question

As the title says, does Boost's cpp_dec_float support modulus operations? I'm dealing with money and would like to do some modulus operations to get the number of bills and coins that I need to return. cpp_dec_float seems to be the only arbitrary precision class that supports base 10.

Answer 1

If all you want is coin distribution, can't you just do it without modulo?

Boost multiprecision does define fmod, trunc, mod etc. whenever possible for a backend type; Here's a working demo: Live On Coliru

#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/number.hpp>
#include <iostream>

using Num = boost::multiprecision::number<boost::multiprecision::cpp_dec_float<128>>;

int main()
{
    Num x("189.5307");

    for (Num denom : { 
            Num("500.00"),   Num("200.00"),   Num("100.00"),
             Num("50.00"),    Num("20.00"),    Num("10.00"),
              Num("5.00"),     Num("2.00"),     Num("1.00"),
              Num("0.50"),     Num("0.20"),     Num("0.10"),
              Num("0.05"),     Num("0.02"),     Num("0.01"),
        })
    {
        Num count = x/denom;

        if (count >= 1)
        {
            count = trunc(count);
            std::cout << count << " * " << std::right << std::setw(6) << denom.str(2, std::ios::fixed) << "\n";
            x -= denom * count;
        }
    }

    std::cout << "Remainder: " << x << "\n";
}

I explicitly selected an "invalid" input value (one with excess precision for the denominations) so you can verify it handles them. I have not bothered with the case for negative amounts, but you can figure that out :)

1 * 100.00
1 *  50.00
1 *  20.00
1 *  10.00
1 *   5.00
2 *   2.00
1 *   0.50
1 *   0.02
1 *   0.01
Remainder: 0.0007