本文介绍了NEON 向量化无符号字节的乘积和:(a[i]-int1) * (b[i]-int2)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我需要改进一个循环,因为它被我的应用程序调用了数千次.我想我需要用 Neon 来做,但我不知道从哪里开始.
I need to improve a loop, because is called by my application thousands of times. I suppose I need to do it with Neon, but I don´t know where to begin.
假设/先决条件:
w
始终为 320(16/32 的倍数).pa
和pb
是 16 字节对齐的ma
和mb
为正.
w
is always 320 (multiple of 16/32).pa
andpb
are 16-byte alignedma
andmb
are positive.
int whileInstruction (const unsigned char *pa,const unsigned char *pb,int ma,int mb,int w)
{
int sum=0;
do {
sum += ((*pa++)-ma)*((*pb++)-mb);
} while(--w);
return sum;
}
这种对其进行矢量化的尝试效果不佳,也不安全(缺少 clobbers),但演示了我正在尝试做的事情:
This attempt at vectorizing it is not working well, and isn't safe (missing clobbers), but demonstrates what I'm trying to do:
int whileInstruction (const unsigned char *pa,const unsigned char *pb,int ma,int mb,int w)
{
asm volatile("lsr %2, %2, #3 \n"
".loop: \n"
"# load 8 elements: \n"
"vld4.8 {d0-d3}, [%1]! \n"
"vld4.8 {d4-d7}, [%2]! \n"
"# do the operation: \n"
"vaddl.u8 q7, d0, r7 \n"
"vaddl.u8 q8, d1, d8 \n"
"vmlal.u8 q7, q7, q8 \n"
"# Sum the vector a save in sum (this is wrong):\n"
"vaddl.u8 q7, d0, r7 \n"
"subs %2, %2, #1 \n" // Decrement iteration count
"bne .loop \n" // Repeat unil iteration count is not zero
:
: "r"(pa), "r"(pb), "r"(w),"r"(ma),"r"(mb),"r"(sum)
: "r4", "r5", "r6","r7","r8","r9"
);
return sum;
}
推荐答案
这是一个简单的 NEON 实现.我已经针对标量代码对此进行了测试,以确保它有效.请注意,为了获得最佳性能,pa
和 pb
都应该是 16 字节对齐的.
Here is a simple NEON implementation. I have tested this against the scalar code to make sure that it works. Note that for best performance both pa
and pb
should be 16 byte aligned.
#include <arm_neon.h>
int whileInstruction_neon(const unsigned char *pa, const unsigned char *pb, int ma, int mb, int w)
{
int sum = 0;
const int32x4_t vma = { ma, ma, ma, ma };
const int32x4_t vmb = { mb, mb, mb, mb };
int32x4_t vsumll = { 0 };
int32x4_t vsumlh = { 0 };
int32x4_t vsumhl = { 0 };
int32x4_t vsumhh = { 0 };
int32x4_t vsum;
int i;
for (i = 0; i <= (w - 16); i += 16)
{
uint8x16_t va = vld1q_u8(pa); // load vector from pa
uint8x16_t vb = vld1q_u8(pb); // load vector from pb
// unpack va into 4 vectors
int16x8_t val = (int16x8_t)vmovl_u8(vget_low_u8(va));
int16x8_t vah = (int16x8_t)vmovl_u8(vget_high_u8(va));
int32x4_t vall = vmovl_s16(vget_low_s16(val));
int32x4_t valh = vmovl_s16(vget_high_s16(val));
int32x4_t vahl = vmovl_s16(vget_low_s16(vah));
int32x4_t vahh = vmovl_s16(vget_high_s16(vah));
// subtract means
vall = vsubq_s32(vall, vma);
valh = vsubq_s32(valh, vma);
vahl = vsubq_s32(vahl, vma);
vahh = vsubq_s32(vahh, vma);
// unpack vb into 4 vectors
int16x8_t vbl = (int16x8_t)vmovl_u8(vget_low_u8(vb));
int16x8_t vbh = (int16x8_t)vmovl_u8(vget_high_u8(vb));
int32x4_t vbll = vmovl_s16(vget_low_s16(vbl));
int32x4_t vblh = vmovl_s16(vget_high_s16(vbl));
int32x4_t vbhl = vmovl_s16(vget_low_s16(vbh));
int32x4_t vbhh = vmovl_s16(vget_high_s16(vbh));
// subtract means
vbll = vsubq_s32(vbll, vmb);
vblh = vsubq_s32(vblh, vmb);
vbhl = vsubq_s32(vbhl, vmb);
vbhh = vsubq_s32(vbhh, vmb);
// update 4 partial sum of products vectors
vsumll = vmlaq_s32(vsumll, vall, vbll);
vsumlh = vmlaq_s32(vsumlh, valh, vblh);
vsumhl = vmlaq_s32(vsumhl, vahl, vbhl);
vsumhh = vmlaq_s32(vsumhh, vahh, vbhh);
pa += 16;
pb += 16;
}
// sum 4 partial sum of product vectors
vsum = vaddq_s32(vsumll, vsumlh);
vsum = vaddq_s32(vsum, vsumhl);
vsum = vaddq_s32(vsum, vsumhh);
// do scalar horizontal sum across final vector
sum = vgetq_lane_s32(vsum, 0);
sum += vgetq_lane_s32(vsum, 1);
sum += vgetq_lane_s32(vsum, 2);
sum += vgetq_lane_s32(vsum, 3);
// handle any residual non-multiple of 16 points
for ( ; i < w; ++i)
{
sum += (*pa++ - ma) * (*pb++ - mb);
}
return sum;
}
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