[Bug libstdc++/77776] C++17 std::hypot implementation is poor

[mkretz at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=77776

--- Comment #15 from Matthias Kretz (Vir) <mkretz at gcc dot gnu.org> ---
Your implementation still needs to solve:

1. Loss of precision because of division & subsequent scaling by max. Users
comparing std::hypot(x, y, z) against a simple std::sqrt(x * x + y * y + z * z)
might wonder why they want to use std::hypot if it's less precise.

2. Relatively high cost (in latency and throughput) because of the three
divisions. You could replace it with scale = 1/max; x *= scale; ... But that
can reduce precision even further.

3. Summation of the x, y, and z squares isn't associative if you care about
precision. A high quality implementation needs to add the two lowest values
first.

Here's a precise but inefficient implementation:
(https://compiler-explorer.com/z/ocGPnsYE3)

template <typename T>
[[gnu::optimize("-fno-unsafe-math-optimizations")]]
T
hypot3(T x, T y, T z)
{
  x = std::abs(x);
  y = std::abs(y);
  z = std::abs(z);
  if (std::isinf(x) || std::isinf(y) || std::isinf(z))
    return std::__infinity_v<T>;
  else if (std::isnan(x) || std::isnan(y) || std::isnan(z))
    return std::__quiet_NaN_v<T>;
  else if (x == y && y == z)
    return x * std::sqrt(T(3));
  else if (z == 0 && y == 0)
    return x;
  else if (x == 0 && z == 0)
    return y;
  else if (x == 0 && y == 0)
    return z;
  else
    {
      T hi = std::max(std::max(x, y), z);
      T lo0 = std::min(std::max(x, y), z);
      T lo1 = std::min(x, y);
      int e = 0;
      hi = std::frexp(hi, &e);
      lo0 = std::ldexp(lo0, -e);
      lo1 = std::ldexp(lo1, -e);
      T lo = lo0 * lo0 + lo1 * lo1;
      return std::ldexp(std::sqrt(hi * hi + lo), e);
    }
}

AFAIK
https://gcc.gnu.org/git/?p=gcc.git;a=blob;f=libstdc%2B%2B-v3/include/experimental/bits/simd_math.h;h=06e7b4496f9917f886f66fbd7629700dd17e55f9;hb=HEAD#l1168
is a precise and efficient implementation. It also avoids division altogether
unless an input is subnormal.