Related
I am trying to understand more about how CPU cache affects performance. As a simple test I am summing the values of the first column of a matrix with varying numbers of total columns.
// compiled with: gcc -Wall -Wextra -Ofast -march=native cache.c
// tested with: for n in {1..100}; do ./a.out $n; done | tee out.csv
#include <assert.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
double sum_column(uint64_t ni, uint64_t nj, double const data[ni][nj])
{
double sum = 0.0;
for (uint64_t i = 0; i < ni; ++i) {
sum += data[i][0];
}
return sum;
}
int compare(void const* _a, void const* _b)
{
double const a = *((double*)_a);
double const b = *((double*)_b);
return (a > b) - (a < b);
}
int main(int argc, char** argv)
{
// set sizes
assert(argc == 2);
uint64_t const iter_max = 101;
uint64_t const ni = 1000000;
uint64_t const nj = strtol(argv[1], 0, 10);
// initialize data
double(*data)[nj] = calloc(ni, sizeof(*data));
for (uint64_t i = 0; i < ni; ++i) {
for (uint64_t j = 0; j < nj; ++j) {
data[i][j] = rand() / (double)RAND_MAX;
}
}
// test performance
double* dt = calloc(iter_max, sizeof(*dt));
double const sum0 = sum_column(ni, nj, data);
for (uint64_t iter = 0; iter < iter_max; ++iter) {
clock_t const t_start = clock();
double const sum = sum_column(ni, nj, data);
clock_t const t_stop = clock();
assert(sum == sum0);
dt[iter] = (t_stop - t_start) / (double)CLOCKS_PER_SEC;
}
// sort dt
qsort(dt, iter_max, sizeof(*dt), compare);
// compute mean dt
double dt_mean = 0.0;
for (uint64_t iter = 0; iter < iter_max; ++iter) {
dt_mean += dt[iter];
}
dt_mean /= iter_max;
// print results
printf("%2lu %.8e %.8e %.8e %.8e\n", nj, dt[iter_max / 2], dt_mean, dt[0],
dt[iter_max - 1]);
// free memory
free(data);
}
However, the results are not quite how I would expect them to be:
As far as I understand, when the CPU loads a value from data, it also places some of the following values of data in the cache. The exact number depends on the cache line size (64 byte on my machine). This would explain, why with growing nj the time to solution first increases linearly and levels out at some value. If nj == 1, one load places the next 7 values in the cache and thus we only need to load from RAM every 8th value. If nj == 2, following the same logic, we need to access the RAM every 4th value. After some size, we will have to access the RAM for every value, which should result in the performance leveling out. My guess, for why the linear section of the graph goes further than 4 is that in reality there are multiple levels of cache at work here and the way that values end up in these caches is a little more complex than what I explained here.
What I cannot explain is why there are these performance peaks at multiples of 16.
After thinking about this question for a bit, I decided to check if this also occurs for higher values of nj:
In fact, it does. And, there is more: Why does the performance increase again after ~250?
Could someone explain to me, or point me to some appropriate reference, why there are these peaks and why the performance increases for higher values of nj.
If you would like to try the code for yourself, I will also attach my plotting script, for your convenience:
import numpy as np
import matplotlib.pyplot as plt
data = np.genfromtxt("out.csv")
data[:,1:] /= data[0,1]
dy = np.diff(data[:,2]) / np.diff(data[:,0])
for i in range(len(dy) - 1):
if dy[i] - dy[i + 1] > (dy.max() - dy.min()) / 2:
plt.axvline(data[i + 1,0], color='gray', linestyle='--')
plt.text(data[i + 1,0], 1.5 * data[0,3], f"{int(data[i + 1,0])}",
rotation=0, ha="center", va="center",
bbox=dict(boxstyle="round", ec='gray', fc='w'))
plt.fill_between(data[:,0], data[:,3], data[:,4], color='gray')
plt.plot(data[:,0], data[:,1], label="median")
plt.plot(data[:,0], data[:,2], label="mean")
plt.legend(loc="upper left")
plt.xlabel("nj")
plt.ylabel("dt / dt$_0$")
plt.savefig("out.pdf")
The plots show the combination of several complex low-level effects (mainly cache trashing & prefetching issues). I assume the target platform is a mainstream modern processor with cache lines of 64 bytes (typically a x86 one).
I can reproduce the problem on my i5-9600KF processor. Here is the resulting plot:
First of all, when nj is small, the gap between fetched address (ie. strides) is small and cache lines are relatively efficiently used. For example, when nj = 1, the access is contiguous. In this case, the processor can efficiently prefetch the cache lines from the DRAM so to hide its high latency. There is also a good spatial cache locality since many contiguous items share the same cache line. When nj=2, only half the value of a cache line is used. This means the number of requested cache line is twice bigger for the same number of operations. That being said the time is not much bigger due to the relatively high latency of adding two floating-point numbers resulting in a compute-bound code. You can unroll the loop 4 times and use 4 different sum variables so that (mainstream modern) processors can add multiple values in parallel. Note that most processors can also load multiple values from the cache per cycle. When nj = 4 a new cache line is requested every 2 cycles (since a double takes 8 bytes). As a result, the memory throughput can become so big that the computation becomes memory-bound. One may expect the time to be stable for nj >= 8 since the number of requested cache line should be the same, but in practice processors prefetch multiple contiguous cache lines so not to pay the overhead of the DRAM latency which is huge in this case. The number of prefetched cache lines is generally between 2 to 4 (AFAIK such prefetching strategy is disabled on Intel processors when the stride is bigger than 512, so when nj >= 64. This explains why the timings are sharply increasing when nj < 32 and they become relatively stable with 32 <= nj <= 256 with exceptions for peaks.
The regular peaks happening when nj is a multiple of 16 are due to a complex cache effect called cache thrashing. Modern cache are N-way associative with N typically between 4 and 16. For example, here are statistics on my i5-9600KF processors:
Cache 0: L1 data cache, line size 64, 8-ways, 64 sets, size 32k
Cache 1: L1 instruction cache, line size 64, 8-ways, 64 sets, size 32k
Cache 2: L2 unified cache, line size 64, 4-ways, 1024 sets, size 256k
Cache 3: L3 unified cache, line size 64, 12-ways, 12288 sets, size 9216k
This means that two fetched values from the DRAM with the respective address A1 and A2 can results in conflicts in my L1 cache if (A1 % 32768) / 64 == (A2 % 32768) / 64. In this case, the processor needs to choose which cache line to replace from a set of N=8 cache lines. There are many cache replacement policy and none is perfect. Thus, some useful cache line are sometime evicted too early resulting in additional cache misses required later. In pathological cases, many DRAM locations can compete for the same cache lines resulting in excessive cache misses. More information about this can be found also in this post.
Regarding the nj stride, the number of cache lines that can be effectively used in the L1 cache is limited. For example, if all fetched values have the same address modulus the cache size, then only N cache lines (ie. 8 for my processor) can actually be used to store all the values. Having less cache lines available is a big problem since the prefetcher need a pretty large space in the cache so to store the many cache lines needed later. The smaller the number of concurrent fetches, the lower memory throughput. This is especially true here since the latency of fetching 1 cache line from the DRAM is about several dozens of nanoseconds (eg. ~70 ns) while its bandwidth is about dozens of GiB/s (eg. ~40 GiB/s): dozens of cache lines (eg. ~40) should be fetched concurrently so to hide the latency and saturate the DRAM.
Here is the simulation of the number of cache lines that can be actually used in my L1 cache regarding the value of the nj:
nj #cache-lines
1 512
2 512
3 512
4 512
5 512
6 512
7 512
8 512
9 512
10 512
11 512
12 512
13 512
14 512
15 512
16 256 <----
17 512
18 512
19 512
20 512
21 512
22 512
23 512
24 512
25 512
26 512
27 512
28 512
29 512
30 512
31 512
32 128 <----
33 512
34 512
35 512
36 512
37 512
38 512
39 512
40 512
41 512
42 512
43 512
44 512
45 512
46 512
47 512
48 256 <----
49 512
50 512
51 512
52 512
53 512
54 512
55 512
56 512
57 512
58 512
59 512
60 512
61 512
62 512
63 512
64 64 <----
==============
80 256
96 128
112 256
128 32
144 256
160 128
176 256
192 64
208 256
224 128
240 256
256 16
384 32
512 8
1024 4
We can see that the number of available cache lines is smaller when nj is a multiple of 16. In this case, the prefetecher will preload data into cache lines that are likely evicted early by subsequent fetched (done concurrently). Loads instruction performed in the code are more likely to result in cache misses when the number of available cache line is small. When a cache miss happen, the value need then to be fetched again from the L2 or even the L3 resulting in a slower execution. Note that the L2 cache is also subject to the same effect though it is less visible since it is larger. The L3 cache of modern x86 processors makes use of hashing to better distributes things to reduce collisions from fixed strides (at least on Intel processors and certainly on AMD too though AFAIK this is not documented).
Here is the timings on my machine for some peaks:
32 4.63600000e-03 4.62298020e-03 4.06400000e-03 4.97300000e-03
48 4.95800000e-03 4.96994059e-03 4.60400000e-03 5.59800000e-03
64 5.01600000e-03 5.00479208e-03 4.26900000e-03 5.33100000e-03
96 4.99300000e-03 5.02284158e-03 4.94700000e-03 5.29700000e-03
128 5.23300000e-03 5.26405941e-03 4.93200000e-03 5.85100000e-03
192 4.76900000e-03 4.78833663e-03 4.60100000e-03 5.01600000e-03
256 5.78500000e-03 5.81666337e-03 5.77600000e-03 6.35300000e-03
384 5.25900000e-03 5.32504950e-03 5.22800000e-03 6.75800000e-03
512 5.02700000e-03 5.05165347e-03 5.02100000e-03 5.34400000e-03
1024 5.29200000e-03 5.33059406e-03 5.28700000e-03 5.65700000e-03
As expected, the timings are overall bigger in practice for the case where the number of available cache lines is much smaller. However, when nj >= 512, the results are surprising since they are significantly faster than others. This is the case where the number of available cache lines is equal to the number of ways of associativity (N). My guess is that this is because Intel processors certainly detect this pathological case and optimize the prefetching so to reduce the number of cache misses (using line-fill buffers to bypass the L1 cache -- see below).
Finally, for large nj stride, a bigger nj should results in higher overheads mainly due to the translation lookaside buffer (TLB): there are more page addresses to translate with bigger nj and the number of TLB entries is limited. In fact this is what I can observe on my machine: timings tends to slowly increase in a very stable way unlike on your target platform.
I cannot really explain this very strange behavior yet.
Here is some wild guesses:
The OS could tend to uses more huge pages when nj is large (so to reduce de overhead of the TLB) since wider blocks are allocated. This could result in more concurrency for the prefetcher as AFAIK it cannot cross page
boundaries. You can try to check the number of allocated (transparent) huge-pages (by looking AnonHugePages in /proc/meminfo in Linux) or force them to be used in this case (using an explicit memmap), or possibly by disabling them. My system appears to make use of 2 MiB transparent huge-pages independently of the nj value.
If the target architecture is a NUMA one (eg. new AMD processors or a server with multiple processors having their own memory), then the OS could allocate pages physically stored on another NUMA node because there is less space available on the current NUMA node. This could result in higher performance due to the bigger throughput (though the latency is higher). You can control this policy with numactl on Linux so to force local allocations.
For more information about this topic, please read the great document What Every Programmer Should Know About Memory. Moreover, a very good post about how x86 cache works in practice is available here.
Removing the peaks
To remove the peaks due to cache trashing on x86 processors, you can use non-temporal software prefetching instructions so cache lines can be fetched in a non-temporal cache structure and into a location close to the processor that should not cause cache trashing in the L1 (if possible). Such cache structure is typically a line-fill buffers (LFB) on Intel processors and the (equivalent) miss address buffers (MAB) on AMD Zen processors. For more information about non-temporal instructions and the LFB, please read this post and this one. Here is the modified code that also include a loop unroling optimization to speed up the code when nj is small:
double sum_column(uint64_t ni, uint64_t nj, double* const data)
{
double sum0 = 0.0;
double sum1 = 0.0;
double sum2 = 0.0;
double sum3 = 0.0;
if(nj % 16 == 0)
{
// Cache-bypassing prefetch to avoid cache trashing
const size_t distance = 12;
for (uint64_t i = 0; i < ni; ++i) {
_mm_prefetch(&data[(i+distance)*nj+0], _MM_HINT_NTA);
sum0 += data[i*nj+0];
}
}
else
{
// Unrolling is much better for small strides
for (uint64_t i = 0; i < ni; i+=4) {
sum0 += data[(i+0)*nj+0];
sum1 += data[(i+1)*nj+0];
sum2 += data[(i+2)*nj+0];
sum3 += data[(i+3)*nj+0];
}
}
return sum0 + sum1 + sum2 + sum3;
}
Here is the result of the modified code:
We can see that peaks no longer appear in the timings. We can also see that the values are much bigger due to dt0 being about 4 times smaller (due to the loop unrolling).
Note that cache trashing in the L2 cache is not avoided with this method in practice (at least on Intel processors). This means that the effect is still here with huge nj strides multiple of 512 (4 KiB) on my machine (it is actually a slower than before, especially when nj >= 2048). It may be a good idea to stop the prefetching when (nj%512) == 0 && nj >= 512 on x86 processors. The effect AFAIK, there is no way to address this problem. That being said, this is a very bad idea to perform such big strided accesses on very-large data structures.
Note that distance should be carefully chosen since early prefetching can result cache line being evicted before they are actually used (so they need to be fetched again) and late prefetching is not much useful. I think using value close to the number of entries in the LFB/MAB is a good idea (eg. 12 on Skylake/KabyLake/CannonLake, 22 on Zen-2).
I am trying to optimize my matrix multiplication code running on a single core. How can I futher improve the performance in regards to loop unrolling, FMA/SSE? I'm also curious to know why the performance won't increase if you use four instead of two sums in the inner loop.
The problem size is a 1000x1000 matrix multiplication. Both gcc 9 and icc 19.0.5 are available. Intel Xeon # 3.10GHz, 32K L1d Cache, Skylake Architecture. Compiled with gcc -O3 -mavx.
void mmult(double* A, double* B, double* C)
{
const int block_size = 64 / sizeof(double);
__m256d sum[2], broadcast;
for (int i0 = 0; i0 < SIZE_M; i0 += block_size) {
for (int k0 = 0; k0 < SIZE_N; k0 += block_size) {
for (int j0 = 0; j0 < SIZE_K; j0 += block_size) {
int imax = i0 + block_size > SIZE_M ? SIZE_M : i0 + block_size;
int kmax = k0 + block_size > SIZE_N ? SIZE_N : k0 + block_size;
int jmax = j0 + block_size > SIZE_K ? SIZE_K : j0 + block_size;
for (int i1 = i0; i1 < imax; i1++) {
for (int k1 = k0; k1 < kmax; k1++) {
broadcast = _mm256_broadcast_sd(A+i1*SIZE_N+k1);
for (int j1 = j0; j1 < jmax; j1+=8) {
sum[0] = _mm256_load_pd(C+i1*SIZE_K+j1+0);
sum[0] = _mm256_add_pd(sum[0], _mm256_mul_pd(broadcast, _mm256_load_pd(B+k1*SIZE_K+j1+0)));
_mm256_store_pd(C+i1*SIZE_K+j1+0, sum[0]);
sum[1] = _mm256_load_pd(C+i1*SIZE_K+j1+4);
sum[1] = _mm256_add_pd(sum[1], _mm256_mul_pd(broadcast, _mm256_load_pd(B+k1*SIZE_K+j1+4)));
_mm256_store_pd(C+i1*SIZE_K+j1+4, sum[1]);
// doesn't improve performance.. why?
// sum[2] = _mm256_load_pd(C+i1*SIZE_K+j1+8);
// sum[2] = _mm256_add_pd(sum[2], _mm256_mul_pd(broadcast, _mm256_load_pd(B+k1*SIZE_K+j1+8)));
// _mm256_store_pd(C+i1*SIZE_K+j1+8, sum[2]);
// sum[3] = _mm256_load_pd(C+i1*SIZE_K+j1+12);
// sum[3] = _mm256_add_pd(sum[3], _mm256_mul_pd(broadcast, _mm256_load_pd(B+k1*SIZE_K+j1+12)));
// _mm256_store_pd(C+i1*SIZE_K+j1+4, sum[3]);
}
}
}
}
}
}
}
This code has 2 loads per FMA (if FMA-contraction happens), but Skylake only supports at most one load per FMA in theory (if you want to max out 2/clock FMA throughput), and even that is usually too much in practice. (Peak through is 2 loads + 1 store per clock, but it usually can't quite sustain that). See Intel's optimization guide and https://agner.org/optimize/
The loop overhead is not the biggest problem, the body itself forces the code to run at half speed.
If the k-loop was the inner loop, a lot of accumulation could be chained, without having to load/store to and from C. This has a downside: with a loop-carried dependency chain like that, it would be up to to code to explicitly ensure that there is enough independent work to be done.
In order to have few loads but enough independent work, the body of the inner loop could calculate the product between a small column vector from A and a small row vector from B, for example using 4 scalar broadcasts to load the column and 2 normal vector loads from B, resulting in just 6 loads for 8 independent FMAs (even lower ratios are possible), which is enough independent FMAs to keep Skylake happy and not too many loads. Even a 3x4 footprint is possible, which also has enough independent FMAs to keep Haswell happy (it needs at least 10).
I happen to have some example code, it's for single precision and C++ but you'll get the point:
sumA_1 = _mm256_load_ps(&result[i * N + j]);
sumB_1 = _mm256_load_ps(&result[i * N + j + 8]);
sumA_2 = _mm256_load_ps(&result[(i + 1) * N + j]);
sumB_2 = _mm256_load_ps(&result[(i + 1) * N + j + 8]);
sumA_3 = _mm256_load_ps(&result[(i + 2) * N + j]);
sumB_3 = _mm256_load_ps(&result[(i + 2) * N + j + 8]);
sumA_4 = _mm256_load_ps(&result[(i + 3) * N + j]);
sumB_4 = _mm256_load_ps(&result[(i + 3) * N + j + 8]);
for (size_t k = kk; k < kk + akb; k++) {
auto bc_mat1_1 = _mm256_set1_ps(*mat1ptr);
auto vecA_mat2 = _mm256_load_ps(mat2 + m2idx);
auto vecB_mat2 = _mm256_load_ps(mat2 + m2idx + 8);
sumA_1 = _mm256_fmadd_ps(bc_mat1_1, vecA_mat2, sumA_1);
sumB_1 = _mm256_fmadd_ps(bc_mat1_1, vecB_mat2, sumB_1);
auto bc_mat1_2 = _mm256_set1_ps(mat1ptr[N]);
sumA_2 = _mm256_fmadd_ps(bc_mat1_2, vecA_mat2, sumA_2);
sumB_2 = _mm256_fmadd_ps(bc_mat1_2, vecB_mat2, sumB_2);
auto bc_mat1_3 = _mm256_set1_ps(mat1ptr[N * 2]);
sumA_3 = _mm256_fmadd_ps(bc_mat1_3, vecA_mat2, sumA_3);
sumB_3 = _mm256_fmadd_ps(bc_mat1_3, vecB_mat2, sumB_3);
auto bc_mat1_4 = _mm256_set1_ps(mat1ptr[N * 3]);
sumA_4 = _mm256_fmadd_ps(bc_mat1_4, vecA_mat2, sumA_4);
sumB_4 = _mm256_fmadd_ps(bc_mat1_4, vecB_mat2, sumB_4);
m2idx += 16;
mat1ptr++;
}
_mm256_store_ps(&result[i * N + j], sumA_1);
_mm256_store_ps(&result[i * N + j + 8], sumB_1);
_mm256_store_ps(&result[(i + 1) * N + j], sumA_2);
_mm256_store_ps(&result[(i + 1) * N + j + 8], sumB_2);
_mm256_store_ps(&result[(i + 2) * N + j], sumA_3);
_mm256_store_ps(&result[(i + 2) * N + j + 8], sumB_3);
_mm256_store_ps(&result[(i + 3) * N + j], sumA_4);
_mm256_store_ps(&result[(i + 3) * N + j + 8], sumB_4);
This means that the j-loop and the i-loop are unrolled, but not the k-loop, even though it is the inner loop now. Unrolling the k-loop a bit did help a bit in my experiments.
See #harold's answer for an actual improvement. This is mostly to repost what I wrote in comments.
four instead of two sums in the inner loop. (Why doesn't unrolling help?)
There's no loop-carried dependency through sum[i]. The next iteration assigns sum[0] = _mm256_load_pd(C+i1*SIZE_K+j1+0); which has no dependency on the previous value.
Therefore register-renaming of the same architectural register onto different physical registers is sufficient to avoid write-after-write hazards that might stall the pipeline. No need to complicate the source with multiple tmp variables. See Why does mulss take only 3 cycles on Haswell, different from Agner's instruction tables? (Unrolling FP loops with multiple accumulators) (In that question, one dot product of 2 arrays, there is a loop carried dependency through an accumulator. There, using multiple accumulators is valuable to hide FP FMA latency so we bottleneck on FMA throughput, not latency.)
A pipeline without register renaming (most in-order CPUs) would benefit from "software pipelining" to statically schedule for what out-of-order exec can do on the fly: load into different registers so there's distance (filled with independent work) between each load and the FMA that consumes it. And then between that and the store.
But all modern x86 CPUs are OoO; even Knight's Landing has some limited OoO exec for SIMD vectors. (Silvermont doesn't support AVX, but does run SIMD instructions in-order, only doing OoO for integer).
Without any multiple-accumulator situation to hide latency, the benefits of unrolling (explicitly in the source or with -funroll-loop as enabled by -fprofile-use, or in clang by default) are:
Reduce front-end bandwidth to get the loop overhead into the back-end. More useful-work uops per loop overhead. Thus it helps if your "useful work" is close to bottlenecked on the front end.
Less back-end execution-unit demand for running the loop overhead. Normally not a problem on Haswell and later, or Zen; the back end can mostly keep up with the front-end when the instruction mix includes some integer stuff and some pure load instructions.
Fewer total uops per work done means OoO exec can "see" farther ahead for memory loads/stores.
Sometimes better branch prediction for short-running loops: The lower iteration count means a shorter pattern for branch prediction to learn. So for short trip-counts, a better chance of correctly predicting the not-taken for the last iteration when execution falls out of the loop.
Sometimes save a mov reg,reg in more complicated cases where it's easier for the compiler to generate a new result in a different reg. The same variable can alternate between living in two regs instead of needing to get moved back to the same one to be ready for the next iteration. Especially if you have a loop that uses a[i] and a[i+1] in a dependent way, or something like Fibonacci.
With 2 loads + 1 store in the loop, that will probably be the bottleneck, not FMA or front-end bandwidth. Unrolling by 2 might have helped avoid a front-end bottleneck, but more than that would only matter with contention from another hyperthread.
An interesting question came up in comments: doesn't unrolling need a lot of registers to be useful?
Harold commented:
16 is not a huge number of registers, but it's enough to have 12
accumulators and 3 pieces of row vector from B and the broadcasted
scalar from A, so it works out to just about enough. The loop from OP
above barely uses any registers anyway. The 8 registers in 32bit are
indeed too few.
Of course since the code in the question doesn't have "accumulators" in registers across loop iterations, only adding into memory, compilers could have optimized all of sum[0..n] to reuse the same register in asm; it's "dead" after storing. So actual register pressure is very low.
Yes x86-64 is somewhat register-poor, that's why AVX512 doubles the number as well as width of vector regs (zmm0..31). Yes, many RISCs have 32 int / 32 fp regs, including AArch64 up from 16 in ARM.
x86-64 has 16 scalar integer registers (including the stack pointer, not including the program counter), so normal functions can use 15. There are also 16 vector regs, xmm0..15. (And with AVX they're double the width ymm0..15).
(Some of this was written before I noticed that sum[0..n] was pointless, not loop-carried.)
Register renaming onto a large physical register file is sufficient in this case. There are other cases where having more architectural registers helps, especially for higher FP latency hence why AVX512 has 32 zmm regs. But for integer 16 is close to enough. RISC CPUs were often designed for in-order without reg renaming, needing SW pipeline.
With OoO exec, the jump from 8 to 16 architectural GP integer regs is more significant than a jump from 16 to 32 would be, in terms of reducing spill/reloads. (I've seen a paper that measured total dynamic instruction count for SPECint with various numbers of architectural registers. I didn't look it up again, but 8->16 might have been 10% total saving while 16->32 was only a couple %. Something like that).
But this specific problem doesn't need a lot of FP registers, only 4 vectors for sum[0..3] (if they were loop-carried) and maybe 1 temporary; x86 can use memory-source mul/add/FMA. Register renaming removes any WAW hazards so we can reuse the same temporary register instead of needing software pipelining. (And OoO exec also hides load and ALU latency.)
You want multiple accumulators when there are loop-carried dependencies. This code is adding into memory, not into a few vector accumulators, so any dependency is through store/reload. But that only has ~7 cycle latency so any sane cache-blocking factor hides it.
I'm trying to enhance the performance of my code by using the 256bit vector (Intel intrinsics - AVX).
I have an I7 Gen.4 (Haswell architecture) processor supporting SSE1 to SSE4.2 and AVX/AVX2 Extensions.
This is the code snippet that I'm trying to enhance:
/* code snipet */
kfac1 = kfac + factor; /* 7 cycles for 7 additions */
kfac2 = kfac1 + factor;
kfac3 = kfac2 + factor;
kfac4 = kfac3 + factor;
kfac5 = kfac4 + factor;
kfac6 = kfac5 + factor;
kfac7 = kfac6 + factor;
k1fac1 = k1fac + factor1; /* 7 cycles for 7 additions */
k1fac2 = k1fac1 + factor1;
k1fac3 = k1fac2 + factor1;
k1fac4 = k1fac3 + factor1;
k1fac5 = k1fac4 + factor1;
k1fac6 = k1fac5 + factor1;
k1fac7 = k1fac6 + factor1;
k2fac1 = k2fac + factor2; /* 7 cycles for 7 additions */
k2fac2 = k2fac1 + factor2;
k2fac3 = k2fac2 + factor2;
k2fac4 = k2fac3 + factor2;
k2fac5 = k2fac4 + factor2;
k2fac6 = k2fac5 + factor2;
k2fac7 = k2fac6 + factor2;
/* code snipet */
From the Intel Manuals, I found this.
an integer addition ADD takes 1 cycle (latency).
a vector of 8 integers (32 bit) takes 1 cycle also.
So I've tried ton make it this way:
fac = _mm256_set1_epi32 (factor )
fac1 = _mm256_set1_epi32 (factor1)
fac2 = _mm256_set1_epi32 (factor2)
v1 = _mm256_set_epi32 (0,kfac6,kfac5,kfac4,kfac3,kfac2,kfac1,kfac)
v2 = _mm256_set_epi32 (0,k1fac6,k1fac5,k1fac4,k1fac3,k1fac2,k1fac1,k1fac)
v3 = _mm256_set_epi32 (0,k2fac6,k2fac5,k2fac4,k2fac3,k2fac2,k2fac1,k2fac)
res1 = _mm256_add_epi32 (v1,fac) ////////////////////
res2 = _mm256_add_epi32 (v2,fa1) // just 3 cycles //
res3 = _mm256_add_epi32 (v3,fa2) ////////////////////
But the problem is that these factors are going to be used as tables indexes ( table[kfac] ... ). So i have to extract the factor as seperate integers again.
I wonder if there is any possible way to do it??
A smart compiler could get table+factor into a register and use indexed addressing modes to get table+factor+k1fac6 as an address. Check the asm, and if the compiler doesn't do this for you, try changing the source to hand-hold the compiler:
const int *tf = table + factor;
const int *tf2 = table + factor2; // could be lea rdx, [rax+rcx*4] or something.
...
foo = tf[kfac2];
bar = tf2[k2fac6]; // could be mov r12, [rdx + rdi*4]
But to answer the question you asked:
Latency isn't a big deal when you have that many independent adds happening. The throughput of 4 scalar add instructions per clock on Haswell is much more relevant.
If k1fac2 and so on are already in contiguous memory, then using SIMD is possibly worth it. Otherwise all the shuffling and data transfer to get them in/out of vector regs makes it definitely not worth it. (i.e. the stuff compiler emits to implement _mm256_set_epi32 (0,kfac6,kfac5,kfac4,kfac3,kfac2,kfac1,kfac).
You could avoid needing to get the indices back into integer registers by using an AVX2 gather for the table loads. But gather is slow on Haswell, so probably not worth it. Maybe worth it on Broadwell.
On Skylake, gather is fast so it could be good if you can SIMD whatever you do with the LUT results. If you need to extract all the gather results back to separate integer registers, it's probably not worth it.
If you did need to extract 8x 32-bit integers from a __m256i into integer registers, you have three main choices of strategy:
Vector store to a tmp array and scalar loads
ALU shuffle instructions like pextrd (_mm_extract_epi32). Use _mm256_extracti128_si256 to get the high lane into a separate __m128i.
A mix of both strategies (e.g. store the high 128 to memory while using ALU stuff on the low half).
Depending on the surrounding code, any of these three could be optimal on Haswell.
pextrd r32, xmm, imm8 is 2 uops on Haswell, with one of them needing the shuffle unit on port5. That's a lot of shuffle uops, so a pure ALU strategy is only going to be good if your code is bottlenecked on L1d cache throughput. (Not the same thing as memory bandwidth). movd r32, xmm is only 1 uop, and compilers do know to use that when compiling _mm_extract_epi32(vec, 0), but you can also write int foo = _mm_cvtsi128_si32(vec) to make it explicit and remind yourself that the bottom element can be accessed more efficiently.
Store/reload has good throughput. Intel SnB-family CPUs including Haswell can run two loads per clock, and IIRC store-forwarding works from an aligned 32-byte store to any 4-byte element of it. But make sure it's an aligned store, e.g. into _Alignas(32) int tmp[8], or into a union between an __m256i and an int array. You could still store into the int array instead of the __m256i member to avoid union type-punning while still having the array aligned, but it's easiest to just use C++11 alignas or C11 _Alignas.
_Alignas(32) int tmp[8];
_mm256_store_si256((__m256i*)tmp, vec);
...
foo2 = tmp[2];
However, the problem with store/reload is latency. Even the first result won't be ready for 6 cycles after the store-data is ready.
A mixed strategy gives you the best of both worlds: ALU to extract the first 2 or 3 elements lets execution get started on whatever code uses them, hiding the store-forwarding latency of the store/reload.
_Alignas(32) int tmp[8];
_mm256_store_si256((__m256i*)tmp, vec);
__m128i lo = _mm256_castsi256_si128(vec); // This is free, no instructions
int foo0 = _mm_cvtsi128_si32(lo);
int foo1 = _mm_extract_epi32(lo, 1);
foo2 = tmp[2];
// rest of foo3..foo7 also loaded from tmp[]
// Then use foo0..foo7
You might find that it's optimal to do the first 4 elements with pextrd, in which case you only need to store/reload the upper lane. Use vextracti128 [mem], ymm, 1:
_Alignas(16) int tmp[4];
_mm_store_si128((__m128i*)tmp, _mm256_extracti128_si256(vec, 1));
// movd / pextrd for foo0..foo3
int foo4 = tmp[0];
...
With fewer larger elements (e.g. 64-bit integers), a pure ALU strategy is more attractive. 6-cycle vector-store / integer-reload latency is longer than it would take to get all of the results with ALU ops, but store/reload could still be good if there's a lot of instruction-level parallelism and you bottleneck on ALU throughput instead of latency.
With more smaller elements (8 or 16-bit), store/reload is definitely attractive. Extracting the first 2 to 4 elements with ALU instructions is still good. And maybe even vmovd r32, xmm and then picking that apart with integer shift/mask instructions is good.
Your cycle-counting for the vector version is also bogus. The three _mm256_add_epi32 operations are independent, and Haswell can run two vpaddd instructions in parallel. (Skylake can run all three in a single cycle, each with 1 cycle latency.)
Superscalar pipelined out-of-order execution means there's a big difference between latency and throughput, and keeping track of dependency chains matters a lot. See http://agner.org/optimize/, and other links in the x86 tag wiki for more optimization guides.
If you have an input array, and an output array, but you only want to write those elements which pass a certain condition, what would be the most efficient way to do this in AVX2?
I've seen in SSE where it was done like this:
(From:https://deplinenoise.files.wordpress.com/2015/03/gdc2015_afredriksson_simd.pdf)
__m128i LeftPack_SSSE3(__m128 mask, __m128 val)
{
// Move 4 sign bits of mask to 4-bit integer value.
int mask = _mm_movemask_ps(mask);
// Select shuffle control data
__m128i shuf_ctrl = _mm_load_si128(&shufmasks[mask]);
// Permute to move valid values to front of SIMD register
__m128i packed = _mm_shuffle_epi8(_mm_castps_si128(val), shuf_ctrl);
return packed;
}
This seems fine for SSE which is 4 wide, and thus only needs a 16 entry LUT, but for AVX which is 8 wide, the LUT becomes quite large(256 entries, each 32 bytes, or 8k).
I'm surprised that AVX doesn't appear to have an instruction for simplifying this process, such as a masked store with packing.
I think with some bit shuffling to count the # of sign bits set to the left you could generate the necessary permutation table, and then call _mm256_permutevar8x32_ps. But this is also quite a few instructions I think..
Does anyone know of any tricks to do this with AVX2? Or what is the most efficient method?
Here is an illustration of the Left Packing Problem from the above document:
Thanks
AVX2 + BMI2. See my other answer for AVX512. (Update: saved a pdep in 64bit builds.)
We can use AVX2 vpermps (_mm256_permutevar8x32_ps) (or the integer equivalent, vpermd) to do a lane-crossing variable-shuffle.
We can generate masks on the fly, since BMI2 pext (Parallel Bits Extract) provides us with a bitwise version of the operation we need.
Beware that pdep/pext are very slow on AMD CPUs before Zen 3, like 6 uops / 18 cycle latency and throughput on Ryzen Zen 1 and Zen 2. This implementation will perform horribly on those AMD CPUs. For AMD, you might be best with 128-bit vectors using a pshufb or vpermilps LUT, or some of the AVX2 variable-shift suggestions discussed in comments. Especially if your mask input is a vector mask (not an already packed bitmask from memory).
AMD before Zen2 only has 128-bit vector execution units anyway, and 256-bit lane-crossing shuffles are slow. So 128-bit vectors are very attractive for this on Zen 1. But Zen 2 has 256-bit load/store and execution units. (And still slow microcoded pext/pdep.)
For integer vectors with 32-bit or wider elements: Either 1) _mm256_movemask_ps(_mm256_castsi256_ps(compare_mask)).
Or 2) use _mm256_movemask_epi8 and then change the first PDEP constant from 0x0101010101010101 to 0x0F0F0F0F0F0F0F0F to scatter blocks of 4 contiguous bits. Change the multiply by 0xFFU into expanded_mask |= expanded_mask<<4; or expanded_mask *= 0x11; (Not tested). Either way, use the shuffle mask with VPERMD instead of VPERMPS.
For 64-bit integer or double elements, everything still Just Works; The compare-mask just happens to always have pairs of 32-bit elements that are the same, so the resulting shuffle puts both halves of each 64-bit element in the right place. (So you still use VPERMPS or VPERMD, because VPERMPD and VPERMQ are only available with immediate control operands.)
For 16-bit elements, you might be able to adapt this with 128-bit vectors.
For 8-bit elements, see Efficient sse shuffle mask generation for left-packing byte elements for a different trick, storing the result in multiple possibly-overlapping chunks.
The algorithm:
Start with a constant of packed 3 bit indices, with each position holding its own index. i.e. [ 7 6 5 4 3 2 1 0 ] where each element is 3 bits wide. 0b111'110'101'...'010'001'000.
Use pext to extract the indices we want into a contiguous sequence at the bottom of an integer register. e.g. if we want indices 0 and 2, our control-mask for pext should be 0b000'...'111'000'111. pext will grab the 010 and 000 index groups that line up with the 1 bits in the selector. The selected groups are packed into the low bits of the output, so the output will be 0b000'...'010'000. (i.e. [ ... 2 0 ])
See the commented code for how to generate the 0b111000111 input for pext from the input vector mask.
Now we're in the same boat as the compressed-LUT: unpack up to 8 packed indices.
By the time you put all the pieces together, there are three total pext/pdeps. I worked backwards from what I wanted, so it's probably easiest to understand it in that direction, too. (i.e. start with the shuffle line, and work backward from there.)
We can simplify the unpacking if we work with indices one per byte instead of in packed 3-bit groups. Since we have 8 indices, this is only possible with 64bit code.
See this and a 32bit-only version on the Godbolt Compiler Explorer. I used #ifdefs so it compiles optimally with -m64 or -m32. gcc wastes some instructions, but clang makes really nice code.
#include <stdint.h>
#include <immintrin.h>
// Uses 64bit pdep / pext to save a step in unpacking.
__m256 compress256(__m256 src, unsigned int mask /* from movmskps */)
{
uint64_t expanded_mask = _pdep_u64(mask, 0x0101010101010101); // unpack each bit to a byte
expanded_mask *= 0xFF; // mask |= mask<<1 | mask<<2 | ... | mask<<7;
// ABC... -> AAAAAAAABBBBBBBBCCCCCCCC...: replicate each bit to fill its byte
const uint64_t identity_indices = 0x0706050403020100; // the identity shuffle for vpermps, packed to one index per byte
uint64_t wanted_indices = _pext_u64(identity_indices, expanded_mask);
__m128i bytevec = _mm_cvtsi64_si128(wanted_indices);
__m256i shufmask = _mm256_cvtepu8_epi32(bytevec);
return _mm256_permutevar8x32_ps(src, shufmask);
}
This compiles to code with no loads from memory, only immediate constants. (See the godbolt link for this and the 32bit version).
# clang 3.7.1 -std=gnu++14 -O3 -march=haswell
mov eax, edi # just to zero extend: goes away when inlining
movabs rcx, 72340172838076673 # The constants are hoisted after inlining into a loop
pdep rax, rax, rcx # ABC -> 0000000A0000000B....
imul rax, rax, 255 # 0000000A0000000B.. -> AAAAAAAABBBBBBBB..
movabs rcx, 506097522914230528
pext rax, rcx, rax
vmovq xmm1, rax
vpmovzxbd ymm1, xmm1 # 3c latency since this is lane-crossing
vpermps ymm0, ymm1, ymm0
ret
(Later clang compiles like GCC, with mov/shl/sub instead of imul, see below.)
So, according to Agner Fog's numbers and https://uops.info/, this is 6 uops (not counting the constants, or the zero-extending mov that disappears when inlined). On Intel Haswell, it's 16c latency (1 for vmovq, 3 for each pdep/imul/pext / vpmovzx / vpermps). There's no instruction-level parallelism. In a loop where this isn't part of a loop-carried dependency, though, (like the one I included in the Godbolt link), the bottleneck is hopefully just throughput, keeping multiple iterations of this in flight at once.
This can maybe manage a throughput of one per 4 cycles, bottlenecked on port1 for pdep/pext/imul plus popcnt in the loop. Of course, with loads/stores and other loop overhead (including the compare and movmsk), total uop throughput can easily be an issue, too.
e.g. the filter loop in my godbolt link is 14 uops with clang, with -fno-unroll-loops to make it easier to read. It might sustain one iteration per 4c, keeping up with the front-end, if we're lucky.
clang 6 and earlier created a loop-carried dependency with popcnt's false dependency on its output, so it will bottleneck on 3/5ths of the latency of the compress256 function. clang 7.0 and later use xor-zeroing to break the false dependency (instead of just using popcnt edx,edx or something like GCC does :/).
gcc (and later clang) does the multiply by 0xFF with multiple instructions, using a left shift by 8 and a sub, instead of imul by 255. This takes 3 total uops vs. 1 for the front-end, but the latency is only 2 cycles, down from 3. (Haswell handles mov at register-rename stage with zero latency.) Most significantly for this, imul can only run on port 1, competing with pdep/pext/popcnt, so it's probably good to avoid that bottleneck.
Since all hardware that supports AVX2 also supports BMI2, there's probably no point providing a version for AVX2 without BMI2.
If you need to do this in a very long loop, the LUT is probably worth it if the initial cache-misses are amortized over enough iterations with the lower overhead of just unpacking the LUT entry. You still need to movmskps, so you can popcnt the mask and use it as a LUT index, but you save a pdep/imul/pext.
You can unpack LUT entries with the same integer sequence I used, but #Froglegs's set1() / vpsrlvd / vpand is probably better when the LUT entry starts in memory and doesn't need to go into integer registers in the first place. (A 32bit broadcast-load doesn't need an ALU uop on Intel CPUs). However, a variable-shift is 3 uops on Haswell (but only 1 on Skylake).
See my other answer for AVX2+BMI2 with no LUT.
Since you mention a concern about scalability to AVX512: don't worry, there's an AVX512F instruction for exactly this:
VCOMPRESSPS — Store Sparse Packed Single-Precision Floating-Point Values into Dense Memory. (There are also versions for double, and 32 or 64bit integer elements (vpcompressq), but not byte or word (16bit)). It's like BMI2 pdep / pext, but for vector elements instead of bits in an integer reg.
The destination can be a vector register or a memory operand, while the source is a vector and a mask register. With a register dest, it can merge or zero the upper bits. With a memory dest, "Only the contiguous vector is written to the destination memory location".
To figure out how far to advance your pointer for the next vector, popcnt the mask.
Let's say you want to filter out everything but values >= 0 from an array:
#include <stdint.h>
#include <immintrin.h>
size_t filter_non_negative(float *__restrict__ dst, const float *__restrict__ src, size_t len) {
const float *endp = src+len;
float *dst_start = dst;
do {
__m512 sv = _mm512_loadu_ps(src);
__mmask16 keep = _mm512_cmp_ps_mask(sv, _mm512_setzero_ps(), _CMP_GE_OQ); // true for src >= 0.0, false for unordered and src < 0.0
_mm512_mask_compressstoreu_ps(dst, keep, sv); // clang is missing this intrinsic, which can't be emulated with a separate store
src += 16;
dst += _mm_popcnt_u64(keep); // popcnt_u64 instead of u32 helps gcc avoid a wasted movsx, but is potentially slower on some CPUs
} while (src < endp);
return dst - dst_start;
}
This compiles (with gcc4.9 or later) to (Godbolt Compiler Explorer):
# Output from gcc6.1, with -O3 -march=haswell -mavx512f. Same with other gcc versions
lea rcx, [rsi+rdx*4] # endp
mov rax, rdi
vpxord zmm1, zmm1, zmm1 # vpxor xmm1, xmm1,xmm1 would save a byte, using VEX instead of EVEX
.L2:
vmovups zmm0, ZMMWORD PTR [rsi]
add rsi, 64
vcmpps k1, zmm0, zmm1, 29 # AVX512 compares have mask regs as a destination
kmovw edx, k1 # There are some insns to add/or/and mask regs, but not popcnt
movzx edx, dx # gcc is dumb and doesn't know that kmovw already zero-extends to fill the destination.
vcompressps ZMMWORD PTR [rax]{k1}, zmm0
popcnt rdx, rdx
## movsx rdx, edx # with _popcnt_u32, gcc is dumb. No casting can get gcc to do anything but sign-extend. You'd expect (unsigned) would mov to zero-extend, but no.
lea rax, [rax+rdx*4] # dst += ...
cmp rcx, rsi
ja .L2
sub rax, rdi
sar rax, 2 # address math -> element count
ret
Performance: 256-bit vectors may be faster on Skylake-X / Cascade Lake
In theory, a loop that loads a bitmap and filters one array into another should run at 1 vector per 3 clocks on SKX / CSLX, regardless of vector width, bottlenecked on port 5. (kmovb/w/d/q k1, eax runs on p5, and vcompressps into memory is 2p5 + a store, according to IACA and to testing by http://uops.info/).
#ZachB reports in comments that in practice, that a loop using ZMM _mm512_mask_compressstoreu_ps is slightly slower than _mm256_mask_compressstoreu_ps on real CSLX hardware. (I'm not sure if that was a microbenchmark that would allow the 256-bit version to get out of "512-bit vector mode" and clock higher, or if there was surrounding 512-bit code.)
I suspect misaligned stores are hurting the 512-bit version. vcompressps probably effectively does a masked 256 or 512-bit vector store, and if that crosses a cache line boundary then it has to do extra work. Since the output pointer is usually not a multiple of 16 elements, a full-line 512-bit store will almost always be misaligned.
Misaligned 512-bit stores may be worse than cache-line-split 256-bit stores for some reason, as well as happening more often; we already know that 512-bit vectorization of other things seems to be more alignment sensitive. That may just be from running out of split-load buffers when they happen every time, or maybe the fallback mechanism for handling cache-line splits is less efficient for 512-bit vectors.
It would be interesting to benchmark vcompressps into a register, with separate full-vector overlapping stores. That's probably the same uops, but the store can micro-fuse when it's a separate instruction. And if there's some difference between masked stores vs. overlapping stores, this would reveal it.
Another idea discussed in comments below was using vpermt2ps to build up full vectors for aligned stores. This would be hard to do branchlessly, and branching when we fill a vector will probably mispredict unless the bitmask has a pretty regular pattern, or big runs of all-0 and all-1.
A branchless implementation with a loop-carried dependency chain of 4 or 6 cycles through the vector being constructed might be possible, with a vpermt2ps and a blend or something to replace it when it's "full". With an aligned vector store every iteration, but only moving the output pointer when the vector is full.
This is likely slower than vcompressps with unaligned stores on current Intel CPUs.
If you are targeting AMD Zen this method may be preferred, due to the very slow pdepand pext on ryzen (18 cycles each).
I came up with this method, which uses a compressed LUT, which is 768(+1 padding) bytes, instead of 8k. It requires a broadcast of a single scalar value, which is then shifted by a different amount in each lane, then masked to the lower 3 bits, which provides a 0-7 LUT.
Here is the intrinsics version, along with code to build LUT.
//Generate Move mask via: _mm256_movemask_ps(_mm256_castsi256_ps(mask)); etc
__m256i MoveMaskToIndices(u32 moveMask) {
u8 *adr = g_pack_left_table_u8x3 + moveMask * 3;
__m256i indices = _mm256_set1_epi32(*reinterpret_cast<u32*>(adr));//lower 24 bits has our LUT
// __m256i m = _mm256_sllv_epi32(indices, _mm256_setr_epi32(29, 26, 23, 20, 17, 14, 11, 8));
//now shift it right to get 3 bits at bottom
//__m256i shufmask = _mm256_srli_epi32(m, 29);
//Simplified version suggested by wim
//shift each lane so desired 3 bits are a bottom
//There is leftover data in the lane, but _mm256_permutevar8x32_ps only examines the first 3 bits so this is ok
__m256i shufmask = _mm256_srlv_epi32 (indices, _mm256_setr_epi32(0, 3, 6, 9, 12, 15, 18, 21));
return shufmask;
}
u32 get_nth_bits(int a) {
u32 out = 0;
int c = 0;
for (int i = 0; i < 8; ++i) {
auto set = (a >> i) & 1;
if (set) {
out |= (i << (c * 3));
c++;
}
}
return out;
}
u8 g_pack_left_table_u8x3[256 * 3 + 1];
void BuildPackMask() {
for (int i = 0; i < 256; ++i) {
*reinterpret_cast<u32*>(&g_pack_left_table_u8x3[i * 3]) = get_nth_bits(i);
}
}
Here is the assembly generated by MSVC:
lea ecx, DWORD PTR [rcx+rcx*2]
lea rax, OFFSET FLAT:unsigned char * g_pack_left_table_u8x3 ; g_pack_left_table_u8x3
vpbroadcastd ymm0, DWORD PTR [rcx+rax]
vpsrlvd ymm0, ymm0, YMMWORD PTR __ymm#00000015000000120000000f0000000c00000009000000060000000300000000
Will add more information to a great answer from #PeterCordes : https://stackoverflow.com/a/36951611/5021064.
I did the implementations of std::remove from C++ standard for integer types with it. The algorithm, once you can do compress, is relatively simple: load a register, compress, store. First I'm going to show the variations and then benchmarks.
I ended up with two meaningful variations on the proposed solution:
__m128i registers, any element type, using _mm_shuffle_epi8 instruction
__m256i registers, element type of at least 4 bytes, using _mm256_permutevar8x32_epi32
When the types are smaller then 4 bytes for 256 bit register, I split them in two 128 bit registers and compress/store each one separately.
Link to compiler explorer where you can see complete assembly (there is a using type and width (in elements per pack) in the bottom, which you can plug in to get different variations) : https://gcc.godbolt.org/z/yQFR2t
NOTE: my code is in C++17 and is using a custom simd wrappers, so I do not know how readable it is. If you want to read my code -> most of it is behind the link in the top include on godbolt. Alternatively, all of the code is on github.
Implementations of #PeterCordes answer for both cases
Note: together with the mask, I also compute the number of elements remaining using popcount. Maybe there is a case where it's not needed, but I have not seen it yet.
Mask for _mm_shuffle_epi8
Write an index for each byte into a half byte: 0xfedcba9876543210
Get pairs of indexes into 8 shorts packed into __m128i
Spread them out using x << 4 | x & 0x0f0f
Example of spreading the indexes. Let's say 7th and 6th elements are picked.
It means that the corresponding short would be: 0x00fe. After << 4 and | we'd get 0x0ffe. And then we clear out the second f.
Complete mask code:
// helper namespace
namespace _compress_mask {
// mmask - result of `_mm_movemask_epi8`,
// `uint16_t` - there are at most 16 bits with values for __m128i.
inline std::pair<__m128i, std::uint8_t> mask128(std::uint16_t mmask) {
const std::uint64_t mmask_expanded = _pdep_u64(mmask, 0x1111111111111111) * 0xf;
const std::uint8_t offset =
static_cast<std::uint8_t>(_mm_popcnt_u32(mmask)); // To compute how many elements were selected
const std::uint64_t compressed_idxes =
_pext_u64(0xfedcba9876543210, mmask_expanded); // Do the #PeterCordes answer
const __m128i as_lower_8byte = _mm_cvtsi64_si128(compressed_idxes); // 0...0|compressed_indexes
const __m128i as_16bit = _mm_cvtepu8_epi16(as_lower_8byte); // From bytes to shorts over the whole register
const __m128i shift_by_4 = _mm_slli_epi16(as_16bit, 4); // x << 4
const __m128i combined = _mm_or_si128(shift_by_4, as_16bit); // | x
const __m128i filter = _mm_set1_epi16(0x0f0f); // 0x0f0f
const __m128i res = _mm_and_si128(combined, filter); // & 0x0f0f
return {res, offset};
}
} // namespace _compress_mask
template <typename T>
std::pair<__m128i, std::uint8_t> compress_mask_for_shuffle_epi8(std::uint32_t mmask) {
auto res = _compress_mask::mask128(mmask);
res.second /= sizeof(T); // bit count to element count
return res;
}
Mask for _mm256_permutevar8x32_epi32
This is almost one for one #PeterCordes solution - the only difference is _pdep_u64 bit (he suggests this as a note).
The mask that I chose is 0x5555'5555'5555'5555. The idea is - I have 32 bits of mmask, 4 bits for each of 8 integers. I have 64 bits that I want to get => I need to convert each bit of 32 bits into 2 => therefore 0101b = 5.The multiplier also changes from 0xff to 3 because I will get 0x55 for each integer, not 1.
Complete mask code:
// helper namespace
namespace _compress_mask {
// mmask - result of _mm256_movemask_epi8
inline std::pair<__m256i, std::uint8_t> mask256_epi32(std::uint32_t mmask) {
const std::uint64_t mmask_expanded = _pdep_u64(mmask, 0x5555'5555'5555'5555) * 3;
const std::uint8_t offset = static_cast<std::uint8_t(_mm_popcnt_u32(mmask)); // To compute how many elements were selected
const std::uint64_t compressed_idxes = _pext_u64(0x0706050403020100, mmask_expanded); // Do the #PeterCordes answer
// Every index was one byte => we need to make them into 4 bytes
const __m128i as_lower_8byte = _mm_cvtsi64_si128(compressed_idxes); // 0000|compressed indexes
const __m256i expanded = _mm256_cvtepu8_epi32(as_lower_8byte); // spread them out
return {expanded, offset};
}
} // namespace _compress_mask
template <typename T>
std::pair<__m256i, std::uint8_t> compress_mask_for_permutevar8x32(std::uint32_t mmask) {
static_assert(sizeof(T) >= 4); // You cannot permute shorts/chars with this.
auto res = _compress_mask::mask256_epi32(mmask);
res.second /= sizeof(T); // bit count to element count
return res;
}
Benchmarks
Processor: Intel Core i7 9700K (a modern consumer level CPU, no AVX-512 support)
Compiler: clang, build from trunk near the version 10 release
Compiler options: --std=c++17 --stdlib=libc++ -g -Werror -Wall -Wextra -Wpedantic -O3 -march=native -mllvm -align-all-functions=7
Micro-benchmarking library: google benchmark
Controlling for code alignment:
If you are not familiar with the concept, read this or watch this
All functions in the benchmark's binary are aligned to 128 byte boundary. Each benchmarking function is duplicated 64 times, with a different noop slide in the beginning of the function (before entering the loop). The main numbers I show is min per each measurement. I think this works since the algorithm is inlined. I'm also validated by the fact that I get very different results. At the very bottom of the answer I show the impact of code alignment.
Note: benchmarking code. BENCH_DECL_ATTRIBUTES is just noinline
Benchmark removes some percentage of 0s from an array. I test arrays with {0, 5, 20, 50, 80, 95, 100} percent of zeroes.
I test 3 sizes: 40 bytes (to see if this is usable for really small arrays), 1000 bytes and 10'000 bytes. I group by size because of SIMD depends on the size of the data and not a number of elements. The element count can be derived from an element size (1000 bytes is 1000 chars but 500 shorts and 250 ints). Since time it takes for non simd code depends mostly on the element count, the wins should be bigger for chars.
Plots: x - percentage of zeroes, y - time in nanoseconds. padding : min indicates that this is minimum among all alignments.
40 bytes worth of data, 40 chars
For 40 bytes this does not make sense even for chars - my implementation gets about 8-10 times slower when using 128 bit registers over non-simd code. So, for example, compiler should be careful doing this.
1000 bytes worth of data, 1000 chars
Apparently the non-simd version is dominated by branch prediction: when we get small amount of zeroes we get a smaller speed up: for no 0s - about 3 times, for 5% zeroes - about 5-6 times speed up. For when the branch predictor can't help the non-simd version - there is about a 27 times speed up. It's an interesting property of simd code that it's performance tends to be much less dependent on of data. Using 128 vs 256 register shows practically no difference, since most of the work is still split into 2 128 registers.
1000 bytes worth of data, 500 shorts
Similar results for shorts except with a much smaller gain - up to 2 times.
I don't know why shorts do that much better than chars for non-simd code: I'd expect shorts to be two times faster, since there are only 500 shorts, but the difference is actually up to 10 times.
1000 bytes worth of data, 250 ints
For a 1000 only 256 bit version makes sense - 20-30% win excluding no 0s to remove what's so ever (perfect branch prediction, no removing for non-simd code).
10'000 bytes worth of data, 10'000 chars
The same order of magnitude wins as as for a 1000 chars: from 2-6 times faster when branch predictor is helpful to 27 times when it's not.
Same plots, only simd versions:
Here we can see about a 10% win from using 256 bit registers and splitting them in 2 128 bit ones: about 10% faster. In size it grows from 88 to 129 instructions, which is not a lot, so might make sense depending on your use-case. For base-line - non-simd version is 79 instructions (as far as I know - these are smaller then SIMD ones though).
10'000 bytes worth of data, 5'000 shorts
From 20% to 9 times win, depending on the data distributions. Not showing the comparison between 256 and 128 bit registers - it's almost the same assembly as for chars and the same win for 256 bit one of about 10%.
10'000 bytes worth of data, 2'500 ints
Seems to make a lot of sense to use 256 bit registers, this version is about 2 times faster compared to 128 bit registers. When comparing with non-simd code - from a 20% win with a perfect branch prediction to 3.5 - 4 times as soon as it's not.
Conclusion: when you have a sufficient amount of data (at least 1000 bytes) this can be a very worthwhile optimisation for a modern processor without AVX-512
PS:
On percentage of elements to remove
On one hand it's uncommon to filter half of your elements. On the other hand a similar algorithm can be used in partition during sorting => that is actually expected to have ~50% branch selection.
Code alignment impact
The question is: how much worth it is, if the code happens to be poorly aligned
(generally speaking - there is very little one can do about it).
I'm only showing for 10'000 bytes.
The plots have two lines for min and for max for each percentage point (meaning - it's not one best/worst code alignment - it's the best code alignment for a given percentage).
Code alignment impact - non-simd
Chars:
From 15-20% for poor branch prediction to 2-3 times when branch prediction helped a lot. (branch predictor is known to be affected by code alignment).
Shorts:
For some reason - the 0 percent is not affected at all. It can be explained by std::remove first doing linear search to find the first element to remove. Apparently linear search for shorts is not affected.
Other then that - from 10% to 1.6-1.8 times worth
Ints:
Same as for shorts - no 0s is not affected. As soon as we go into remove part it goes from 1.3 times to 5 times worth then the best case alignment.
Code alignment impact - simd versions
Not showing shorts and ints 128, since it's almost the same assembly as for chars
Chars - 128 bit register
About 1.2 times slower
Chars - 256 bit register
About 1.1 - 1.24 times slower
Ints - 256 bit register
1.25 - 1.35 times slower
We can see that for simd version of the algorithm, code alignment has significantly less impact compared to non-simd version. I suspect that this is due to practically not having branches.
In case anyone is interested here is a solution for SSE2 which uses an instruction LUT instead of a data LUT aka a jump table. With AVX this would need 256 cases though.
Each time you call LeftPack_SSE2 below it uses essentially three instructions: jmp, shufps, jmp. Five of the sixteen cases don't need to modify the vector.
static inline __m128 LeftPack_SSE2(__m128 val, int mask) {
switch(mask) {
case 0:
case 1: return val;
case 2: return _mm_shuffle_ps(val,val,0x01);
case 3: return val;
case 4: return _mm_shuffle_ps(val,val,0x02);
case 5: return _mm_shuffle_ps(val,val,0x08);
case 6: return _mm_shuffle_ps(val,val,0x09);
case 7: return val;
case 8: return _mm_shuffle_ps(val,val,0x03);
case 9: return _mm_shuffle_ps(val,val,0x0c);
case 10: return _mm_shuffle_ps(val,val,0x0d);
case 11: return _mm_shuffle_ps(val,val,0x34);
case 12: return _mm_shuffle_ps(val,val,0x0e);
case 13: return _mm_shuffle_ps(val,val,0x38);
case 14: return _mm_shuffle_ps(val,val,0x39);
case 15: return val;
}
}
__m128 foo(__m128 val, __m128 maskv) {
int mask = _mm_movemask_ps(maskv);
return LeftPack_SSE2(val, mask);
}
This is perhaps a bit late though I recently ran into this exact problem and found an alternative solution which used a strictly AVX implementation. If you don't care if unpacked elements are swapped with the last elements of each vector, this could work as well. The following is an AVX version:
inline __m128 left_pack(__m128 val, __m128i mask) noexcept
{
const __m128i shiftMask0 = _mm_shuffle_epi32(mask, 0xA4);
const __m128i shiftMask1 = _mm_shuffle_epi32(mask, 0x54);
const __m128i shiftMask2 = _mm_shuffle_epi32(mask, 0x00);
__m128 v = val;
v = _mm_blendv_ps(_mm_permute_ps(v, 0xF9), v, shiftMask0);
v = _mm_blendv_ps(_mm_permute_ps(v, 0xF9), v, shiftMask1);
v = _mm_blendv_ps(_mm_permute_ps(v, 0xF9), v, shiftMask2);
return v;
}
Essentially, each element in val is shifted once to the left using the bitfield, 0xF9 for blending with it's unshifted variant. Next, both shifted and unshifted versions are blended against the input mask (which has the first non-zero element broadcast across the remaining elements 3 and 4). Repeat this process two more times, broadcasting the second and third elements of mask to its subsequent elements on each iteration and this should provide an AVX version of the _pdep_u32() BMI2 instruction.
If you don't have AVX, you can easily swap out each _mm_permute_ps() with _mm_shuffle_ps() for an SSE4.1-compatible version.
And if you're using double-precision, here's an additional version for AVX2:
inline __m256 left_pack(__m256d val, __m256i mask) noexcept
{
const __m256i shiftMask0 = _mm256_permute4x64_epi64(mask, 0xA4);
const __m256i shiftMask1 = _mm256_permute4x64_epi64(mask, 0x54);
const __m256i shiftMask2 = _mm256_permute4x64_epi64(mask, 0x00);
__m256d v = val;
v = _mm256_blendv_pd(_mm256_permute4x64_pd(v, 0xF9), v, shiftMask0);
v = _mm256_blendv_pd(_mm256_permute4x64_pd(v, 0xF9), v, shiftMask1);
v = _mm256_blendv_pd(_mm256_permute4x64_pd(v, 0xF9), v, shiftMask2);
return v;
}
Additionally _mm_popcount_u32(_mm_movemask_ps(val)) can be used to determine the number of elements which remained after the left-packing.
Can anyone give an example or a link to an example which uses __builtin_prefetch in GCC (or just the asm instruction prefetcht0 in general) to gain a substantial performance advantage? In particular, I'd like the example to meet the following criteria:
It is a simple, small, self-contained example.
Removing the __builtin_prefetch instruction results in performance degradation.
Replacing the __builtin_prefetch instruction with the corresponding memory access results in performance degradation.
That is, I want the shortest example showing __builtin_prefetch performing an optimization that couldn't be managed without it.
Here's an actual piece of code that I've pulled out of a larger project. (Sorry, it's the shortest one I can find that had a noticable speedup from prefetching.)
This code performs a very large data transpose.
This example uses the SSE prefetch instructions, which may be the same as the one that GCC emits.
To run this example, you will need to compile this for x64 and have more than 4GB of memory. You can run it with a smaller datasize, but it will be too fast to time.
#include <iostream>
using std::cout;
using std::endl;
#include <emmintrin.h>
#include <malloc.h>
#include <time.h>
#include <string.h>
#define ENABLE_PREFETCH
#define f_vector __m128d
#define i_ptr size_t
inline void swap_block(f_vector *A,f_vector *B,i_ptr L){
// To be super-optimized later.
f_vector *stop = A + L;
do{
f_vector tmpA = *A;
f_vector tmpB = *B;
*A++ = tmpB;
*B++ = tmpA;
}while (A < stop);
}
void transpose_even(f_vector *T,i_ptr block,i_ptr x){
// Transposes T.
// T contains x columns and x rows.
// Each unit is of size (block * sizeof(f_vector)) bytes.
//Conditions:
// - 0 < block
// - 1 < x
i_ptr row_size = block * x;
i_ptr iter_size = row_size + block;
// End of entire matrix.
f_vector *stop_T = T + row_size * x;
f_vector *end = stop_T - row_size;
// Iterate each row.
f_vector *y_iter = T;
do{
// Iterate each column.
f_vector *ptr_x = y_iter + block;
f_vector *ptr_y = y_iter + row_size;
do{
#ifdef ENABLE_PREFETCH
_mm_prefetch((char*)(ptr_y + row_size),_MM_HINT_T0);
#endif
swap_block(ptr_x,ptr_y,block);
ptr_x += block;
ptr_y += row_size;
}while (ptr_y < stop_T);
y_iter += iter_size;
}while (y_iter < end);
}
int main(){
i_ptr dimension = 4096;
i_ptr block = 16;
i_ptr words = block * dimension * dimension;
i_ptr bytes = words * sizeof(f_vector);
cout << "bytes = " << bytes << endl;
// system("pause");
f_vector *T = (f_vector*)_mm_malloc(bytes,16);
if (T == NULL){
cout << "Memory Allocation Failure" << endl;
system("pause");
exit(1);
}
memset(T,0,bytes);
// Perform in-place data transpose
cout << "Starting Data Transpose... ";
clock_t start = clock();
transpose_even(T,block,dimension);
clock_t end = clock();
cout << "Done" << endl;
cout << "Time: " << (double)(end - start) / CLOCKS_PER_SEC << " seconds" << endl;
_mm_free(T);
system("pause");
}
When I run it with ENABLE_PREFETCH enabled, this is the output:
bytes = 4294967296
Starting Data Transpose... Done
Time: 0.725 seconds
Press any key to continue . . .
When I run it with ENABLE_PREFETCH disabled, this is the output:
bytes = 4294967296
Starting Data Transpose... Done
Time: 0.822 seconds
Press any key to continue . . .
So there's a 13% speedup from prefetching.
EDIT:
Here's some more results:
Operating System: Windows 7 Professional/Ultimate
Compiler: Visual Studio 2010 SP1
Compile Mode: x64 Release
Intel Core i7 860 # 2.8 GHz, 8 GB DDR3 # 1333 MHz
Prefetch : 0.868
No Prefetch: 0.960
Intel Core i7 920 # 3.5 GHz, 12 GB DDR3 # 1333 MHz
Prefetch : 0.725
No Prefetch: 0.822
Intel Core i7 2600K # 4.6 GHz, 16 GB DDR3 # 1333 MHz
Prefetch : 0.718
No Prefetch: 0.796
2 x Intel Xeon X5482 # 3.2 GHz, 64 GB DDR2 # 800 MHz
Prefetch : 2.273
No Prefetch: 2.666
Binary search is a simple example that could benefit from explicit prefetching. The access pattern in a binary search looks pretty much random to the hardware prefetcher, so there is little chance that it will accurately predict what to fetch.
In this example, I prefetch the two possible 'middle' locations of the next loop iteration in the current iteration. One of the prefetches will probably never be used, but the other will (unless this is the final iteration).
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
int binarySearch(int *array, int number_of_elements, int key) {
int low = 0, high = number_of_elements-1, mid;
while(low <= high) {
mid = (low + high)/2;
#ifdef DO_PREFETCH
// low path
__builtin_prefetch (&array[(mid + 1 + high)/2], 0, 1);
// high path
__builtin_prefetch (&array[(low + mid - 1)/2], 0, 1);
#endif
if(array[mid] < key)
low = mid + 1;
else if(array[mid] == key)
return mid;
else if(array[mid] > key)
high = mid-1;
}
return -1;
}
int main() {
int SIZE = 1024*1024*512;
int *array = malloc(SIZE*sizeof(int));
for (int i=0;i<SIZE;i++){
array[i] = i;
}
int NUM_LOOKUPS = 1024*1024*8;
srand(time(NULL));
int *lookups = malloc(NUM_LOOKUPS * sizeof(int));
for (int i=0;i<NUM_LOOKUPS;i++){
lookups[i] = rand() % SIZE;
}
for (int i=0;i<NUM_LOOKUPS;i++){
int result = binarySearch(array, SIZE, lookups[i]);
}
free(array);
free(lookups);
}
When I compile and run this example with DO_PREFETCH enabled, I see a 20% reduction in runtime:
$ gcc c-binarysearch.c -DDO_PREFETCH -o with-prefetch -std=c11 -O3
$ gcc c-binarysearch.c -o no-prefetch -std=c11 -O3
$ perf stat -e L1-dcache-load-misses,L1-dcache-loads ./with-prefetch
Performance counter stats for './with-prefetch':
356,675,702 L1-dcache-load-misses # 41.39% of all L1-dcache hits
861,807,382 L1-dcache-loads
8.787467487 seconds time elapsed
$ perf stat -e L1-dcache-load-misses,L1-dcache-loads ./no-prefetch
Performance counter stats for './no-prefetch':
382,423,177 L1-dcache-load-misses # 97.36% of all L1-dcache hits
392,799,791 L1-dcache-loads
11.376439030 seconds time elapsed
Notice that we are doing twice as many L1 cache loads in the prefetch version. We're actually doing a lot more work but the memory access pattern is more friendly to the pipeline. This also shows the tradeoff. While this block of code runs faster in isolation, we have loaded a lot of junk into the caches and this may put more pressure on other parts of the application.
I learned a lot from the excellent answers provided by #JamesScriven and #Mystical. However, their examples give only a modest boost - the objective of this answer is to present a (I must confess somewhat artificial) example, where prefetching has a bigger impact (about factor 4 on my machine).
There are three possible bottle-necks for the modern architectures: CPU-speed, memory-band-width and memory latency. Prefetching is all about reducing the latency of the memory-accesses.
In a perfect scenario, where latency corresponds to X calculation-steps, we would have a oracle, which would tell us which memory we would access in X calculation-steps, the prefetching of this data would be launched and it would arrive just in-time X calculation-steps later.
For a lot of algorithms we are (almost) in this perfect world. For a simple for-loop it is easy to predict which data will be needed X steps later. Out-of-order execution and other hardware tricks are doing a very good job here, concealing the latency almost completely.
That is the reason, why there is such a modest improvement for #Mystical's example: The prefetcher is already pretty good - there is just not much room for improvement. The task is also memory-bound, so probably not much band-width is left - it could be becoming the limiting factor. I could see at best around 8% improvement on my machine.
The crucial insight from the #JamesScriven example: neither we nor the CPU knows the next access-address before the the current data is fetched from memory - this dependency is pretty important, otherwise out-of-order execution would lead to a look-forward and the hardware would be able to prefetch the data. However, because we can speculate about only one step there is not that much potential. I was not able to get more than 40% on my machine.
So let's rig the competition and prepare the data in such a way that we know which address is accessed in X steps, but make it impossible for hardware to find it out due to dependencies on not yet accessed data (see the whole program at the end of the answer):
//making random accesses to memory:
unsigned int next(unsigned int current){
return (current*10001+328)%SIZE;
}
//the actual work is happening here
void operator()(){
//set up the oracle - let see it in the future oracle_offset steps
unsigned int prefetch_index=0;
for(int i=0;i<oracle_offset;i++)
prefetch_index=next(prefetch_index);
unsigned int index=0;
for(int i=0;i<STEP_CNT;i++){
//use oracle and prefetch memory block used in a future iteration
if(prefetch){
__builtin_prefetch(mem.data()+prefetch_index,0,1);
}
//actual work, the less the better
result+=mem[index];
//prepare next iteration
prefetch_index=next(prefetch_index); #update oracle
index=next(mem[index]); #dependency on `mem[index]` is VERY important to prevent hardware from predicting future
}
}
Some remarks:
data is prepared in such a way, that the oracle is alway right.
maybe surprisingly, the less CPU-bound task the bigger the speed-up: we are able to hide the latency almost completely, thus the speed-up is CPU-time+original-latency-time/CPU-time.
Compiling and executing leads:
>>> g++ -std=c++11 prefetch_demo.cpp -O3 -o prefetch_demo
>>> ./prefetch_demo
#preloops time no prefetch time prefetch factor
...
7 1.0711102260000001 0.230566831 4.6455521002498408
8 1.0511602149999999 0.22651144600000001 4.6406494398521474
9 1.049024333 0.22841439299999999 4.5926367389641687
....
to a speed-up between 4 and 5.
Listing of prefetch_demp.cpp:
//prefetch_demo.cpp
#include <vector>
#include <iostream>
#include <iomanip>
#include <chrono>
const int SIZE=1024*1024*1;
const int STEP_CNT=1024*1024*10;
unsigned int next(unsigned int current){
return (current*10001+328)%SIZE;
}
template<bool prefetch>
struct Worker{
std::vector<int> mem;
double result;
int oracle_offset;
void operator()(){
unsigned int prefetch_index=0;
for(int i=0;i<oracle_offset;i++)
prefetch_index=next(prefetch_index);
unsigned int index=0;
for(int i=0;i<STEP_CNT;i++){
//prefetch memory block used in a future iteration
if(prefetch){
__builtin_prefetch(mem.data()+prefetch_index,0,1);
}
//actual work:
result+=mem[index];
//prepare next iteration
prefetch_index=next(prefetch_index);
index=next(mem[index]);
}
}
Worker(std::vector<int> &mem_):
mem(mem_), result(0.0), oracle_offset(0)
{}
};
template <typename Worker>
double timeit(Worker &worker){
auto begin = std::chrono::high_resolution_clock::now();
worker();
auto end = std::chrono::high_resolution_clock::now();
return std::chrono::duration_cast<std::chrono::nanoseconds>(end-begin).count()/1e9;
}
int main() {
//set up the data in special way!
std::vector<int> keys(SIZE);
for (int i=0;i<SIZE;i++){
keys[i] = i;
}
Worker<false> without_prefetch(keys);
Worker<true> with_prefetch(keys);
std::cout<<"#preloops\ttime no prefetch\ttime prefetch\tfactor\n";
std::cout<<std::setprecision(17);
for(int i=0;i<20;i++){
//let oracle see i steps in the future:
without_prefetch.oracle_offset=i;
with_prefetch.oracle_offset=i;
//calculate:
double time_with_prefetch=timeit(with_prefetch);
double time_no_prefetch=timeit(without_prefetch);
std::cout<<i<<"\t"
<<time_no_prefetch<<"\t"
<<time_with_prefetch<<"\t"
<<(time_no_prefetch/time_with_prefetch)<<"\n";
}
}
From the documentation:
for (i = 0; i < n; i++)
{
a[i] = a[i] + b[i];
__builtin_prefetch (&a[i+j], 1, 1);
__builtin_prefetch (&b[i+j], 0, 1);
/* ... */
}
Pre-fetching data can be optimized to the Cache Line size, which for most modern 64-bit processors is 64 bytes to for example pre-load a uint32_t[16] with one instruction.
For example on ArmV8 I discovered through experimentation casting the memory pointer to a uint32_t 4x4 matrix vector (which is 64 bytes in size) halved the required instructions required as before I had to increment by 8 as it was only loading half the data, even though my understanding was that it fetches a full cache line.
Pre-fetching an uint32_t[32] original code example...
int addrindex = &B[0];
__builtin_prefetch(&V[addrindex]);
__builtin_prefetch(&V[addrindex + 8]);
__builtin_prefetch(&V[addrindex + 16]);
__builtin_prefetch(&V[addrindex + 24]);
After...
int addrindex = &B[0];
__builtin_prefetch((uint32x4x4_t *) &V[addrindex]);
__builtin_prefetch((uint32x4x4_t *) &V[addrindex + 16]);
For some reason int datatype for the address index/offset gave better performance. Tested with GCC 8 on Cortex-a53. Using an equivalent 64 byte vector on other architectures might give the same performance improvement if you find it is not pre-fetching all the data like in my case. In my application with a one million iteration loop, it improved performance by 5% just by doing this. There were further requirements for the improvement.
the 128 megabyte "V" memory allocation had to be aligned to 64 bytes.
uint32_t *V __attribute__((__aligned__(64))) = (uint32_t *)(((uintptr_t)(__builtin_assume_aligned((unsigned char*)aligned_alloc(64,size), 64)) + 63) & ~ (uintptr_t)(63));
Also, I had to use C operators instead of Neon Intrinsics, since they require regular datatype pointers (in my case it was uint32_t *) otherwise the new built in prefetch method had a performance regression.
My real world example can be found at https://github.com/rollmeister/veriumMiner/blob/main/algo/scrypt.c in the scrypt_core() and its internal function which are all easy to read. The hard work is done by GCC8. Overall improvement to performance was 25%.