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Force cpp_dec_float to round down

I am using .str(n, std::ios_base::scientific) to print ccp_dec_floats.

I've noticed that it rounds up.

I am using cpp_dec_float for accounting, so I need to round downward. How can this be done?


1 Answers

It doesn't round up. In fact, it does banker's round: See it Live On Coliru

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

namespace mp = boost::multiprecision;

int main()
{
    using Dec = mp::cpp_dec_float_50;

    for (Dec d : { 
            Dec( "3.34"),   Dec( "3.35"),   Dec( "3.38"),
            Dec( "2.24"),   Dec( "2.25"),   Dec( "2.28"),
            Dec("-2.24"),   Dec("-2.25"),   Dec("-2.28"),
            Dec("-3.34"),   Dec("-3.35"),   Dec("-3.38"),
            })
    {
        std::cout     << d.str(2, std::ios_base::fixed) 
            << " -> " << d.str(1, std::ios_base::fixed) << "\n";
    }
}

Prints:

3.34 -> 3.3
3.35 -> 3.4
3.38 -> 3.4
2.24 -> 2.2
2.25 -> 2.2
2.28 -> 2.3
-2.24 -> -2.2
-2.25 -> -2.2
-2.28 -> -2.3
-3.34 -> -3.3
-3.35 -> -3.4
-3.38 -> -3.4

So if you want another kind of rounding, you'd want to write it explicitly

Here's a generic approach (Live On Coliru)

template <int decimals = 0, typename T>
T round_towards_zero(T const& v)
{
    static const T scale = pow(T(10), decimals);

    if (v.is_zero())
        return v;

    // ceil/floor is found via ADL and uses expression templates for optimization
    if (v<0)
        return ceil(v*scale)/scale;
    else
        // floor is found via ADL and uses expression templates for optimization
        return floor(v*scale)/scale;
}

which hopefully compiles down to optimal code due to statically known scale factor and the use of expression templates in Boost Multiprecision library.

like image 165
sehe Avatar answered Feb 09 '26 09:02

sehe



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