x86 - Sum reduction of unsigned bytes without overflow, using SSE2 on Intel

Question

I am trying to find sum reduction of 32 elements (each 1 byte data) on an Intel i3 processor. I did this:

s=0; 
for (i=0; i<32; i++)
{
    s = s + a[i];
}  

However, its taking more time, since my application is a real-time application requiring much lesser time. Please note that the final sum could be more than 255.

Is there a way I can implement this using low level SIMD SSE2 instructions? Unfortunately I have never used SSE. I tried searching for sse2 function for this purpose, but it is also not available. Is it (sse) guaranteed to reduce the computation time for such a small-sized problems?

Any suggestions??

Note: I have implemented the similar algorithms using OpenCL and CUDA and that worked great but only when the problem size was big. For small sized problems the cost of overhead was more. Not sure how it works on SSE

Answer 1

You can abuse PSADBW to calculate small horizontal sums quickly.

Something like this: (not tested)

pxor xmm0, xmm0
psadbw xmm0, [a + 0]
pxor xmm1, xmm1
psadbw xmm1, [a + 16]
paddw xmm0, xmm1
pshufd xmm1, xmm0, 2
paddw xmm0, xmm1 ; low word in xmm0 is the total sum

---

Attempted intrinsics version:

I never use intrinsics so this code probably makes no sense whatsoever. The disassembly looked OK though.

uint16_t sum_32(const uint8_t a[32])
{
    __m128i zero = _mm_xor_si128(zero, zero);
    __m128i sum0 = _mm_sad_epu8(
                        zero,
                        _mm_load_si128(reinterpret_cast<const __m128i*>(a)));
    __m128i sum1 = _mm_sad_epu8(
                        zero,
                        _mm_load_si128(reinterpret_cast<const __m128i*>(&a[16])));
    __m128i sum2 = _mm_add_epi16(sum0, sum1);
    __m128i totalsum = _mm_add_epi16(sum2, _mm_shuffle_epi32(sum2, 2));
    return totalsum.m128i_u16[0];
}