ternary operator for clang's extended vectors

Question

I\'ve tried playing with clang\'s extended vectors. The ternary operator is supposed to work, but it is not working for me. Example:

int main()
{
  using int4 =          


        

Answer 1

In the end I went with this:

#if defined(__clang__)
template <typename U, typename V>
constexpr inline std::enable_if_t<
  !std::is_arithmetic<V>{},
  V
>
select(V const a, V const b, U const c) noexcept
{
  return V((c & U(a)) | (~c & U(b)));
}
#else
template <typename U, typename V>
constexpr inline std::enable_if_t<
  !std::is_arithmetic<V>{},
  V
>
select(V const a, V const b, U const c) noexcept
{
  return c ? a : b;
}
#endif

The same could have been accomplished in other ways, using the indices trick, for example, but it might not optimize very well (I didn't want any conditionals in there).

Answer 2

You can loop over the elements directly in Clang. Here is a solution for GCC and Clang.

#include <inttypes.h>
#include <x86intrin.h>

#if defined(__clang__)
typedef float float4 __attribute__ ((ext_vector_type(4)));
typedef   int   int4 __attribute__ ((ext_vector_type(4)));
#else
typedef float float4 __attribute__ ((vector_size (sizeof(float)*4)));
typedef   int   int4 __attribute__ ((vector_size (sizeof(int)*4)));
#endif

float4 select(int4 s, float4 a, float4 b) {
  float4 c;
  #if defined(__GNUC__) && !defined(__INTEL_COMPILER) && !defined(__clang__)
  c = s ? a : b;
  #else
  for(int i=0; i<4; i++) c[i] = s[i] ? a[i] : b[i];
  #endif
  return c;
}

The both generate

select(int __vector(4), float __vector(4), float __vector(4)):
  pxor xmm3, xmm3
  pcmpeqd xmm0, xmm3
  blendvps xmm1, xmm2, xmm0
  movaps xmm0, xmm1
  ret

• Nehalem: https://godbolt.org/g/cVWYym
• Skylake: https://godbolt.org/g/LhEpnN
• KNL: https://godbolt.org/g/NFrFKg

But with AVX512 it's better to use masks (e.g. __mmask16).

Answer 3

This works in a pinch:

auto const diff = a-b;
auto const ra( - (diff!=zero) * a - (diff==zero) *b);

I guess this is a bug in the compiler, or in the documentation you linked.