我想使用 SIMD(SSE2 等)优化以下功能:
int64_t fun(int64_t N, int size, int* p)
{
int64_t sum = 0;
for(int i=1; i<size; i++)
sum += (N/i)*p[i];
return sum;
}
这似乎是一个非常可矢量化的任务,只是所需的指令不存在......
我们可以假设 N 非常大(10^12 到 10^18)并且大小~sqrt(N)。我们还可以假设p只能取-1、0和1的值;所以我们不需要真正的乘法,如果我们能以某种方式计算 N/i,则 (N/i)*p[i] 可以用四个指令(pcmpgt、pxor、psub、pand)来完成。
最佳答案
这是我能得到的最接近向量化该代码的结果。我真的不希望它会更快。我只是尝试编写 SIMD 代码。
#include <stdint.h>
int64_t fun(int64_t N, int size, const int* p)
{
int64_t sum = 0;
int i;
for(i=1; i<size; i++) {
sum += (N/i)*p[i];
}
return sum;
}
typedef int64_t v2sl __attribute__ ((vector_size (2*sizeof(int64_t))));
int64_t fun_simd(int64_t N, int size, const int* p)
{
int64_t sum = 0;
int i;
v2sl v_2 = { 2, 2 };
v2sl v_N = { N, N };
v2sl v_i = { 1, 2 };
union { v2sl v; int64_t a[2]; } v_sum;
v_sum.a[0] = 0;
v_sum.a[1] = 0;
for(i=1; i<size-1; i+=2) {
v2sl v_p = { p[i], p[i+1] };
v_sum.v += (v_N / v_i) * v_p;
v_i += v_2;
}
sum = v_sum.a[0] + v_sum.a[1];
for(; i<size; i++) {
sum += (N/i)*p[i];
}
return sum;
}
typedef double v2df __attribute__ ((vector_size (2*sizeof(double))));
int64_t fun_simd_double(int64_t N, int size, const int* p)
{
int64_t sum = 0;
int i;
v2df v_2 = { 2, 2 };
v2df v_N = { N, N };
v2df v_i = { 1, 2 };
union { v2df v; double a[2]; } v_sum;
v_sum.a[0] = 0;
v_sum.a[1] = 0;
for(i=1; i<size-1; i+=2) {
v2df v_p = { p[i], p[i+1] };
v_sum.v += (v_N / v_i) * v_p;
v_i += v_2;
}
sum = v_sum.a[0] + v_sum.a[1];
for(; i<size; i++) {
sum += (N/i)*p[i];
}
return sum;
}
#include <stdio.h>
static const int test_array[] = {
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0,
1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0
};
#define test_array_len (sizeof(test_array)/sizeof(int))
#define big_N (1024 * 1024 * 1024)
int main(int argc, char *argv[]) {
int64_t res1;
int64_t res2;
int64_t res3;
v2sl a = { 123, 456 };
v2sl b = { 100, 200 };
union { v2sl v; int64_t a[2]; } tmp;
a = a + b;
tmp.v = a;
printf("a = { %ld, %ld }\n", tmp.a[0], tmp.a[1]);
printf("test_array size = %zd\n", test_array_len);
res1 = fun(big_N, test_array_len, test_array);
printf("fun() = %ld\n", res1);
res2 = fun_simd(big_N, test_array_len, test_array);
printf("fun_simd() = %ld\n", res2);
res3 = fun_simd_double(big_N, test_array_len, test_array);
printf("fun_simd_double() = %ld\n", res3);
return 0;
}
关于algorithm - SIMD 优化难题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4047319/