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];
}         


        

Answer 1

There is one more way to find the sum of all elements of an array using SSE instructions. The code uses the following SSE constructs.

• __m256 register
• _mm256_store_ps(float *a, __m256 b)
• _mm256_add_ps(__m256 a, __m256 b)

The code works for any sized array of floats.

float sse_array_sum(float *a, int size)
{
    /*
     *   sum += a[i] (for all i in domain)
     */

    float *sse_sum, sum=0;
    if(size >= 8)
    {
        // sse_sum[8]
        posix_memalign((void **)&sse_sum, 32, 8*sizeof(float));

        __m256 temp_sum;
        __m256* ptr_a = (__m256*)a;
        int itrs = size/8-1;

        // sse_sum[0:7] = a[0:7]
        temp_sum = *ptr_a;
        a += 8;
        ptr_a++;

        for(int i=0; i

Benchmark:

size = 64000000
a[i] = 3141592.65358 for all i in domain

sequential version time: 194ms
SSE version time: 49ms

Machine specification:

Thread(s) per core: 2
Core(s) per socket: 2
Socket(s): 1
CPU MHz: 1700.072
OS: Ubuntu