Division that rounds down rather than truncates

Question

I need to align real values divided by some positive constant to nearest lower integer, regardless of their sign. Example are ( here the backslash represents my desired operator)

21,5 \ 2 = 10
-21,5 \ 2 = -11
52,3 \ 2 = 26
-52,3 \ 2 = -27

Is there a short operator that does this ? the usual "slash" ( "/" ) operator rounds towards zero in C++ (that was made standard some time ago) (e.g. -52.6 / 2 = -26).

Float division doesn't do any rounding at all. `-52.6/2` is `-26.30000000000000071054273576010018587112427` on my machine. — molbdnilo, Sep 21 '16 at 09:40
@Leon This only works if a is a float. I sometimes have integers, indifferently. — Charles, Sep 21 '16 at 09:41

score 2 · Answer 1 · answered Sep 21 '16 at 09:46

std::floor will solve your problem.

#include <iostream>
#include <cmath>
using namespace std;

int main() {
    // your code goes here
    float i = -21.5,b=2;
    int c = std::floor(i/b);
    cout << c << endl;

    i = 21.5,b=2;
    c = std::floor(i/b);
    cout << c << endl;

    int a = 11,b1 =2;
    c = std::floor(a/b1);
    cout << c << endl;

    a = -11;
    b =2.1;
    c = std::floor(a/b);
    cout << c << endl;
    return 0;
}

Output:

score 0 · Answer 2 · answered Sep 21 '16 at 09:46

We don't have a special operator for it, but we could make a special type, and redefine appropriate operators:

#include <iostream>
#include <cmath>

template<class Integer>
struct RoundDown
{
    RoundDown(Integer v) : value_(v) {}

    operator Integer&() { return value_; }
    operator const Integer&() const { return value_; }

    template<class Other>
    RoundDown& operator/=(Other const& o)
    {
        value_ = Integer(std::floor(double(value_) / o));
        return *this;
    }

    Integer value_;
};

template<class IntegerL, class IntegerR>
auto operator/(RoundDown<IntegerL> l, IntegerR const& r)
{
    return l /= r;
}

int main()
{
    RoundDown<int> a { -57 };
    a /= 2;
    std::cout << a << '\n';
    std::cout << (a / 2) << '\n';
}

Expected output:

-29
-15

OP wanted to divide real numbers by integers, not integers by integers. I think a function would be more intuitive for that than a class. — anatolyg, Sep 21 '16 at 10:07
@anatolyg oddly enough, even though the template argument is called `Integer`, this class will work for real numbers too. However, I'm sure you're right about intuitiveness. — Richard Hodges, Sep 21 '16 at 10:14

Division that rounds down rather than truncates

2 Answers2