Boost cpp_dec_float supports modulus operation - c++

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