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62 lines
1.8 KiB
C
62 lines
1.8 KiB
C
/* Compute the dot product of two (properly aligned) vectors. */
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#include <stdio.h>
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#include <immintrin.h>
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#include <math.h>
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const int N = 83;
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double slow_dot_product(const double *a, const double *b) {
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double answer = 0.0;
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for(int ii = 0; ii < N; ++ii)
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answer += a[ii]*b[ii];
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return answer;
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}
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/* Horizontal add works within 128-bit lanes. Use scalar ops to add
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* across the boundary. */
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double reduce_vector1(__m256d input) {
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__m256d temp = _mm256_hadd_pd(input, input);
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return ((double*)&temp)[0] + ((double*)&temp)[2];
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}
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/* Another way to get around the 128-bit boundary: grab the first 128
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* bits, grab the lower 128 bits and then add them together with a 128
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* bit add instruction. */
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double reduce_vector2(__m256d input) {
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__m256d temp = _mm256_hadd_pd(input, input);
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__m128d sum_high = _mm256_extractf128_pd(temp, 1);
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__m128d result = _mm_add_pd(sum_high, _mm256_castpd256_pd128(temp));
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return ((double*)&result)[0];
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}
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double dot_product(const double *a, const double *b) {
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__m256d sum_vec = _mm256_set_pd(0.0, 0.0, 0.0, 0.0);
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/* Add up partial dot-products in blocks of 256 bits */
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for(int ii = 0; ii < N/4; ++ii) {
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__m256d x = _mm256_load_pd(a+4*ii);
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__m256d y = _mm256_load_pd(b+4*ii);
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__m256d z = _mm256_mul_pd(x,y);
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sum_vec = _mm256_add_pd(sum_vec, z);
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}
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/* Find the partial dot-product for the remaining elements after
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* dealing with all 256-bit blocks. */
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double final = 0.0;
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for(int ii = N-N%4; ii < N; ++ii)
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final += a[ii] * b[ii];
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return reduce_vector2(sum_vec) + final;
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}
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int main() {
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__attribute__ ((aligned (32))) double a[N], b[N];
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for(int ii = 0; ii < N; ++ii)
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a[ii] = b[ii] = ii/sqrt(N);
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double answer = dot_product(a, b);
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printf("%f\n", answer);
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printf("%f\n", slow_dot_product(a,b));
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}
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