c++visual-studio-2012intrinsicsavxmsvc12

unresolved external symbol __mm256_setr_epi64x


I have written and debugged some AVX code with g++ and now I'm trying to get it to work with MSVC, but I keep getting

error LNK2019: unresolved external symbol __mm256_setr_epi64x referenced in function "private: union __m256i __thiscall avx_matrix::avx_bit_mask(unsigned int)const " (?avx_bit_mask@avx_matrix@@ABE?AT__m256i@@I@Z)

The referenced piece of code is

...

#include <immintrin.h>

...

    /* All zeros except for pos-th position (0..255) */
    __m256i avx_matrix::avx_bit_mask(const std::size_t pos) const
    {
        int64_t a = (pos >= 0 && pos < 64) ? 1LL << (pos - 0) : 0;
        int64_t b = (pos >= 64 && pos < 128) ? 1LL << (pos - 64) : 0;
        int64_t c = (pos >= 128 && pos < 192) ? 1LL << (pos - 128) : 0;
        int64_t d = (pos >= 192 && pos < 256) ? 1LL << (pos - 256) : 0;
        return _mm256_setr_epi64x(a, b, c, d);
    }
...

Any help would be much appreciated.


Solution

  • It looks this might actually be a known bug - certain AVX intrinsics are apparently not available in 32-bit mode. Try building for 64 bit and/or upgrading to Visual Studio 2013 Update 2, where this has supposedly now been fixed.

    Alternatively, if you just have the one instance above where you are using this intrinsic, then you could change your function to:

    __m256i avx_matrix::avx_bit_mask(const std::size_t pos) const
    {
        int64_t a[4] = { (pos >=   0 && pos <  64) ? 1LL << (pos -   0) : 0,
                         (pos >=  64 && pos < 128) ? 1LL << (pos -  64) : 0,
                         (pos >= 128 && pos < 192) ? 1LL << (pos - 128) : 0,
                         (pos >= 192 && pos < 256) ? 1LL << (pos - 256) : 0 };
        return _mm256_loadu_si256((__m256i *)a);
    }
    

    or perhaps even:

    __m256i avx_matrix::avx_bit_mask(const std::size_t pos) const
    {
        int64_t a[4] = { 0 };
        a[pos >> 6] = 1LL << (pos & 63ULL);
        return _mm256_loadu_si256((__m256i *)a);
    }
    

    which might be a little more efficient.