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better sqrt_impl

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MUCH bigger range (although still not quitre max -- see comments)
no recursion
make this example possible: https://godbolt.org/z/tDFkfR
This commit is contained in:
Oliver Schönrock
2020-02-23 15:37:32 +00:00
committed by Mateusz Pusz
parent d4492524d1
commit fe998e1eab
1 changed files with 18 additions and 9 deletions
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27
src/include/units/ratio.h
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@@ -205,18 +205,27 @@ using ratio_pow = detail::ratio_pow_impl<R, N>::type;
namespace detail { namespace detail {
constexpr std::intmax_t sqrt_impl(std::intmax_t v, std::intmax_t l, std::intmax_t r) // sqrt_impl avoids overflow and recursion
// from http://www.codecodex.com/wiki/Calculate_an_integer_square_root#C.2B.2B
// if v >= place this will fail (so we can't quite use the last bit)
static constexpr std::intmax_t sqrt_impl(std::intmax_t v)
{ {
if (l == r) return r; // place = 0x4000 0000 for 32bit
// place = 0x4000 0000 0000 0000 for 64bit
std::intmax_t place = static_cast<std::intmax_t>(1) << (sizeof(std::intmax_t) * 8 - 2);
while (place > v) place /= 4; // optimized by complier as place >>= 2
const auto mid = (r + l) / 2; std::intmax_t root = 0;
if (mid * mid >= v) while (place) {
return sqrt_impl(v, l, mid); if (v >= root + place) {
else v -= root + place;
return sqrt_impl(v, mid + 1, r); root += place * 2;
}
root /= 2;
place /= 4;
}
return root;
} }
static constexpr std::intmax_t sqrt_impl(std::intmax_t v) { return sqrt_impl(v, 1, v); }
template<Ratio R> template<Ratio R>
constexpr auto make_exp_even() constexpr auto make_exp_even()