通过四舍五入提升 rational_cast？

Question

如何进行 rational_cast<int64_t> 舍入？

目前我正在做这样的黑客攻击：

boost::rational<int64_t> pts = ..., time_base = ...;
int64_t rounded = std::llround(boost::rational_cast<long double>(pts / time_base));

但我希望能够在不涉及浮点的情况下做到这一点"properly"。

Answer 1

舍入本质上是有损的。

想到的最快的破解方法就是使用内置行为（即 floor-ing 或 trunc-ing 结果）并偏移一半：

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#include <iostream>
#include <fstream>
#include <boost/rational.hpp>

int main() {
    using R = boost::rational<int64_t>;
    for (auto den : {5,6}) {
        std::cout << "---------\n";
        for (auto num : {1,2,3,4,5,6}) {
            R pq(num, den);
            std::cout << num << "/" << den << " = " << pq << ": " 
                      << boost::rational_cast<int64_t>(pq + R(1,2)) << "\n";
        }
    }
}

版画

---------
1/5 = 1/5: 0
2/5 = 2/5: 0
3/5 = 3/5: 1
4/5 = 4/5: 1
5/5 = 1/1: 1
6/5 = 6/5: 1
---------
1/6 = 1/6: 0
2/6 = 1/3: 0
3/6 = 1/2: 1
4/6 = 2/3: 1
5/6 = 5/6: 1
6/6 = 1/1: 1

通过四舍五入提升 rational_cast？

boost rational_cast with rounding?

c++

rational-number

boost

rounding