I am trying to speed up a single program by using prefetches. The purpose of my program is just for test. Here is what it does:
It uses two int buffers of the same size
It reads one-by-one all the values of the first buffer
It reads the value at the index in the second buffer
It sums all the values taken from the second buffer
It does all the previous steps for bigger and bigger
At the end, I print the number of voluntary and involuntary CPU
In the very first time, values in the first buffers contains the values of its index (cf. function createIndexBuffer in the code just below) .
It will be more clear in the code of my program:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <sys/time.h>
#define BUFFER_SIZE ((unsigned long) 4096 * 100000)
unsigned int randomUint()
{
int value = rand() % UINT_MAX;
return value;
}
unsigned int * createValueBuffer()
{
unsigned int * valueBuffer = (unsigned int *) malloc(BUFFER_SIZE * sizeof(unsigned int));
for (unsigned long i = 0 ; i < BUFFER_SIZE ; i++)
{
valueBuffer[i] = randomUint();
}
return (valueBuffer);
}
unsigned int * createIndexBuffer()
{
unsigned int * indexBuffer = (unsigned int *) malloc(BUFFER_SIZE * sizeof(unsigned int));
for (unsigned long i = 0 ; i < BUFFER_SIZE ; i++)
{
indexBuffer[i] = i;
}
return (indexBuffer);
}
unsigned long long computeSum(unsigned int * indexBuffer, unsigned int * valueBuffer)
{
unsigned long long sum = 0;
for (unsigned int i = 0 ; i < BUFFER_SIZE ; i++)
{
unsigned int index = indexBuffer[i];
sum += valueBuffer[index];
}
return (sum);
}
unsigned int computeTimeInMicroSeconds()
{
unsigned int * valueBuffer = createValueBuffer();
unsigned int * indexBuffer = createIndexBuffer();
struct timeval startTime, endTime;
gettimeofday(&startTime, NULL);
unsigned long long sum = computeSum(indexBuffer, valueBuffer);
gettimeofday(&endTime, NULL);
printf("Sum = %llu\n", sum);
free(indexBuffer);
free(valueBuffer);
return ((endTime.tv_sec - startTime.tv_sec) * 1000 * 1000) + (endTime.tv_usec - startTime.tv_usec);
}
int main()
{
printf("sizeof buffers = %ldMb\n", BUFFER_SIZE * sizeof(unsigned int) / (1024 * 1024));
unsigned int timeInMicroSeconds = computeTimeInMicroSeconds();
printf("Time: %u micro-seconds = %.3f seconds\n", timeInMicroSeconds, (double) timeInMicroSeconds / (1000 * 1000));
}
If I launch it, I get the following output:
$ gcc TestPrefetch.c -O3 -o TestPrefetch && ./TestPrefetch
sizeof buffers = 1562Mb
Sum = 439813150288855829
Time: 201172 micro-seconds = 0.201 seconds
Quick and fast!!!
According to my knowledge (I may be wrong), one of the reason for having such a fast program is that, as I access my two buffers sequentially, data can be prefetched in the CPU cache.
We can make it more complex in order that data is (almost) prefeched in CPU cache. For example, we can just change the createIndexBuffer function in:
unsigned int * createIndexBuffer()
{
unsigned int * indexBuffer = (unsigned int *) malloc(BUFFER_SIZE * sizeof(unsigned int));
for (unsigned long i = 0 ; i < BUFFER_SIZE ; i++)
{
indexBuffer[i] = rand() % BUFFER_SIZE;
}
return (indexBuffer);
}
Let's try the program once again:
$ gcc TestPrefetch.c -O3 -o TestPrefetch && ./TestPrefetch
sizeof buffers = 1562Mb
Sum = 439835307963131237
Time: 3730387 micro-seconds = 3.730 seconds
More than 18 times slower!!!
We now arrive to my problem. Given the new createIndexBuffer function, I would like to speed up computeSum function using prefetch
unsigned long long computeSum(unsigned int * indexBuffer, unsigned int * valueBuffer)
{
unsigned long long sum = 0;
for (unsigned int i = 0 ; i < BUFFER_SIZE ; i++)
{
__builtin_prefetch((char *) &indexBuffer[i + 1], 0, 0);
unsigned int index = indexBuffer[i];
sum += valueBuffer[index];
}
return (sum);
}
of course I also have to change my createIndexBuffer in order it allocates a buffer having one more element
I relaunch my program: not better! As prefetch may be slower than one "for" loop iteration, I may prefetch not one element before but two elements before
__builtin_prefetch((char *) &indexBuffer[i + 2], 0, 0);
not better! two loops iterations? not better? Three? **I tried it until 50 (!!!) but I cannot enhance the performance of my function computeSum.
Can I would like help to understand why
Thank you very much for your help
I believe that above code is automatically optimized by CPU without any further space for manual optimization.
1. Main problem is that indexBuffer is sequentially accessed. Hardware prefetcher senses it and prefetches further values automatically, without need to call prefetch manually. So, during iteration #i, values indexBuffer[i+1], indexBuffer[i+2],... are already in cache. (By the way, there is no need to add artificial element to the end of array: memory access errors are silently ignored by prefetch instructions).
What you really need to do is to prefetch valueBuffer instead:
__builtin_prefetch((char *) &valueBuffer[indexBuffer[i + 1]], 0, 0);
2. But adding above line of code won't help either in such simple scenario. Cost of accessing memory is hundreds of cycles, while add instruction is ~1 cycle. Your code already spends 99% of time in memory accesses. Adding manual prefetch will make it this one cycle faster and no better.
Manual prefetch would really work well if your math were much more heavy (try it), like using an expression with large number of non-optimized out divisions (20-30 cycles each) or calling some math function (log, sin).
3. But even this doesn't guarantee to help. Dependency between loop iterations is very weak, it is only via sum variable. This allows CPU to execute instructions speculatively: it may start fetching valueBuffer[i+1] concurrently while still executing math for valueBuffer[i].
Prefetch fetches normally a full cache line. This is typically 64 bytes. So the random example fetches always 64 bytes for a 4 byte int. 16 times the data you actually need which fits very well with the slow down by a factor of 18. So the code is simply limited by memory throughput and not latency.
Sorry. What I gave you was not the correct version of my code. The correct version is, what you said:
__builtin_prefetch((char *) &valueBuffer[indexBuffer[i + prefetchStep]], 0, 0);
However, even with the right version, it is unfortunately not better
Then I adapted my program to try your suggestion using the sin function.
My adapted program is the following one:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <sys/time.h>
#include <math.h>
#define BUFFER_SIZE ((unsigned long) 4096 * 50000)
unsigned int randomUint()
{
int value = rand() % UINT_MAX;
return value;
}
unsigned int * createValueBuffer()
{
unsigned int * valueBuffer = (unsigned int *) malloc(BUFFER_SIZE * sizeof(unsigned int));
for (unsigned long i = 0 ; i < BUFFER_SIZE ; i++)
{
valueBuffer[i] = randomUint();
}
return (valueBuffer);
}
unsigned int * createIndexBuffer(unsigned short prefetchStep)
{
unsigned int * indexBuffer = (unsigned int *) malloc((BUFFER_SIZE + prefetchStep) * sizeof(unsigned int));
for (unsigned long i = 0 ; i < BUFFER_SIZE ; i++)
{
indexBuffer[i] = rand() % BUFFER_SIZE;
}
return (indexBuffer);
}
double computeSum(unsigned int * indexBuffer, unsigned int * valueBuffer, unsigned short prefetchStep)
{
double sum = 0;
for (unsigned int i = 0 ; i < BUFFER_SIZE ; i++)
{
__builtin_prefetch((char *) &valueBuffer[indexBuffer[i + prefetchStep]], 0, 0);
unsigned int index = indexBuffer[i];
sum += sin(valueBuffer[index]);
}
return (sum);
}
unsigned int computeTimeInMicroSeconds(unsigned short prefetchStep)
{
unsigned int * valueBuffer = createValueBuffer();
unsigned int * indexBuffer = createIndexBuffer(prefetchStep);
struct timeval startTime, endTime;
gettimeofday(&startTime, NULL);
double sum = computeSum(indexBuffer, valueBuffer, prefetchStep);
gettimeofday(&endTime, NULL);
printf("prefetchStep = %d, Sum = %f - ", prefetchStep, sum);
free(indexBuffer);
free(valueBuffer);
return ((endTime.tv_sec - startTime.tv_sec) * 1000 * 1000) + (endTime.tv_usec - startTime.tv_usec);
}
int main()
{
printf("sizeof buffers = %ldMb\n", BUFFER_SIZE * sizeof(unsigned int) / (1024 * 1024));
for (unsigned short prefetchStep = 0 ; prefetchStep < 250 ; prefetchStep++)
{
unsigned int timeInMicroSeconds = computeTimeInMicroSeconds(prefetchStep);
printf("Time: %u micro-seconds = %.3f seconds\n", timeInMicroSeconds, (double) timeInMicroSeconds / (1000 * 1000));
}
}
The output is:
$ gcc TestPrefetch.c -O3 -o TestPrefetch -lm && taskset -c 7 ./TestPrefetch
sizeof buffers = 781Mb
prefetchStep = 0, Sum = -1107.523504 - Time: 20895326 micro-seconds = 20.895 seconds
prefetchStep = 1, Sum = 13456.262424 - Time: 12706720 micro-seconds = 12.707 seconds
prefetchStep = 2, Sum = -20179.289469 - Time: 12136174 micro-seconds = 12.136 seconds
prefetchStep = 3, Sum = 12068.302534 - Time: 11233803 micro-seconds = 11.234 seconds
prefetchStep = 4, Sum = 21071.238160 - Time: 10855348 micro-seconds = 10.855 seconds
prefetchStep = 5, Sum = -22648.280105 - Time: 10517861 micro-seconds = 10.518 seconds
prefetchStep = 6, Sum = 22665.381676 - Time: 9205809 micro-seconds = 9.206 seconds
prefetchStep = 7, Sum = 2461.741268 - Time: 11391088 micro-seconds = 11.391 seconds
...
So here, it works better! Honestly, I was almost sure that it will not be better because the math function cost is higher compared to the memory access.
If anyone could give me more information about why it is better now, I would appreciate it
Thank you very much
Related
Hello I have seen that C++ Vector vs Array (Time).
On my mac the vector take times to be defined but after the comparison give vector for winner.
How it works ?
I was said int[] are faster than dynamic vector ?
#include <iostream>
#include <vector>
using namespace std;
#define N (100000000)
//int sd[N];
int main() {
clock_t start;
double temps;
static int sd[N];
start = clock();
for (unsigned long i=0 ; i < N ; i++){
if(sd[i]==3)
;
}
temps = (clock() - start) / (double)(CLOCKS_PER_SEC / 1000);
printf("Time: %f ms\n",temps);
vector<int>vd(N);
start = clock();
for (unsigned long i=0 ; i < N ; i++){
if(vd[i]==3)
;
}
temps = (clock() - start) / (double)(CLOCKS_PER_SEC / 1000);
printf("Time: %f ms\n",temps);
while (1)
;
return 0;
}
I have those results :
Time: 422.87400 ms
Time: 300.84700 ms
Even if it begining with vector, vector appear to be faster than c array.
Thank You for your explaination.
Another question : in xcode, why i see memory used by declation vector and for static c array I have to go all the memory cells as in the code (for ... if(sd[i]...)
Thank You for your explaination.
I have remarqued that if i initialize all the c array cells at 0 (for example or 6...) the c array will be faster or equal vector.
int main() {
clock_t start;
double temps;
static int sd[N];
for (unsigned long i=0 ; i < N ; i++){
sd[i]=0;
}
start = clock();
//puts("initialized");
for (unsigned long i=0 ; i < N ; i++){
if(sd[i]==3)
;
}
temps = (clock() - start) / (double)(CLOCKS_PER_SEC / 1000);
printf("Time: %f ms\n",temps);
//puts("initialized");
vector<int>vd(N);
start = clock();
for (unsigned long i=0 ; i < N ; i++){
if(vd[i]==3)
;
}
temps = (clock() - start) / (double)(CLOCKS_PER_SEC / 1000);
printf("Time: %f ms\n",temps);
while (1)
;
return 0;
}
And I will see the memory used in xcode with the initialization at 0 of all cells c array.
So another question , why it is more rapid when you initialize in this case (or in general) ?
Summary:
Any ideas about how to further improve upon the basic scatter operation in CUDA? Especially if one knows it will only be used to compact a larger array into a smaller one? or why the below methods of vectorizing memory ops and shared memory didn't work? I feel like there may be something fundamental I am missing and any help would be appreciated.
EDIT 03/09/15: So I found this Parallel For All Blog post "Optimized Filtering with Warp-Aggregated Atomics". I had assumed atomics would be intrinsically slower for this purpose, however I was wrong - especially since I don't think I care about maintaining element order in the array during my simulation. I'll have to think about it some more and then implement it to see what happens!
EDIT 01/04/16: I realized I never wrote about my results. Unfortunately in that Parallel for All Blog post they compared the global atomic method for compact to the Thrust prefix-sum compact method, which is actually quite slow. CUB's Device::IF is much faster than Thrust's - as is the prefix-sum version I wrote using CUB's Device::Scan + custom code. The warp-aggregrate global atomic method is still faster by about 5-10%, but nowhere near the 3-4x faster I had been hoping for based on the results in the blog. I'm still using the prefix-sum method as while maintaining element order is not necessary, I prefer the consistency of the prefix-sum results and the advantage from the atomics is not very big. I still try various methods to improve compact, but so far only marginal improvements (2%) at best for dramatically increased code complexity.
Details:
I am writing a simulation in CUDA where I compact out elements I am no longer interested in simulating every 40-60 time steps. From profiling it seems that the scatter op takes up the most amount of time when compacting - more so than the filter kernel or the prefix sum. Right now I use a pretty basic scatter function:
__global__ void scatter_arrays(float * new_freq, const float * const freq, const int * const flag, const int * const scan_Index, const int freq_Index){
int myID = blockIdx.x*blockDim.x + threadIdx.x;
for(int id = myID; id < freq_Index; id+= blockDim.x*gridDim.x){
if(flag[id]){
new_freq[scan_Index[id]] = freq[id];
}
}
}
freq_Index is the number of elements in the old array. The flag array is the result from the filter. Scan_ID is the result from the prefix sum on the flag array.
Attempts I've made to improve it are to read the flagged frequencies into shared memory first and then write from shared memory to global memory - the idea being that the writes to global memory would be more coalesced amongst the warps (e.g. instead of thread 0 writing to position 0 and thread 128 writing to position 1, thread 0 would write to 0 and thread 1 would write to 1). I also tried vectorizing the reads and the writes - instead of reading and writing floats/ints I read/wrote float4/int4 from the global arrays when possible, so four numbers at a time. This I thought might speed up the scatter by having fewer memory ops transferring larger amounts of memory. The "kitchen sink" code with both vectorized memory loads/stores and shared memory is below:
const int compact_threads = 256;
__global__ void scatter_arrays2(float * new_freq, const float * const freq, const int * const flag, const int * const scan_Index, const int freq_Index){
int gID = blockIdx.x*blockDim.x + threadIdx.x; //global ID
int tID = threadIdx.x; //thread ID within block
__shared__ float row[4*compact_threads];
__shared__ int start_index[1];
__shared__ int end_index[1];
float4 myResult;
int st_index;
int4 myFlag;
int4 index;
for(int id = gID; id < freq_Index/4; id+= blockDim.x*gridDim.x){
if(tID == 0){
index = reinterpret_cast<const int4*>(scan_Index)[id];
myFlag = reinterpret_cast<const int4*>(flag)[id];
start_index[0] = index.x;
st_index = index.x;
myResult = reinterpret_cast<const float4*>(freq)[id];
if(myFlag.x){ row[0] = myResult.x; }
if(myFlag.y){ row[index.y-st_index] = myResult.y; }
if(myFlag.z){ row[index.z-st_index] = myResult.z; }
if(myFlag.w){ row[index.w-st_index] = myResult.w; }
}
__syncthreads();
if(tID > 0){
myFlag = reinterpret_cast<const int4*>(flag)[id];
st_index = start_index[0];
index = reinterpret_cast<const int4*>(scan_Index)[id];
myResult = reinterpret_cast<const float4*>(freq)[id];
if(myFlag.x){ row[index.x-st_index] = myResult.x; }
if(myFlag.y){ row[index.y-st_index] = myResult.y; }
if(myFlag.z){ row[index.z-st_index] = myResult.z; }
if(myFlag.w){ row[index.w-st_index] = myResult.w; }
if(tID == blockDim.x -1 || gID == mutations_Index/4 - 1){ end_index[0] = index.w + myFlag.w; }
}
__syncthreads();
int count = end_index[0] - st_index;
int rem = st_index & 0x3; //equivalent to modulo 4
int offset = 0;
if(rem){ offset = 4 - rem; }
if(tID < offset && tID < count){
new_mutations_freq[population*new_array_Length+st_index+tID] = row[tID];
}
int tempID = 4*tID+offset;
if((tempID+3) < count){
reinterpret_cast<float4*>(new_freq)[tID] = make_float4(row[tempID],row[tempID+1],row[tempID+2],row[tempID+3]);
}
tempID = tID + offset + (count-offset)/4*4;
if(tempID < count){ new_freq[st_index+tempID] = row[tempID]; }
}
int id = gID + freq_Index/4 * 4;
if(id < freq_Index){
if(flag[id]){
new_freq[scan_Index[id]] = freq[id];
}
}
}
Obviously it gets a bit more complicated. :) While the above kernel seems stable when there are hundreds of thousands of elements in the array, I've noticed a race condition when the array numbers in the tens of millions. I'm still trying to track the bug down.
But regardless, neither method (shared memory or vectorization) together or alone improved performance. I was especially surprised by the lack of benefit from vectorizing the memory ops. It had helped in other functions I had written, though now I am wondering if maybe it helped because it increased Instruction-Level-Parallelism in the calculation steps of those other functions rather than the fewer memory ops.
I found the algorithm mentioned in this poster (similar algorithm also discussed in this paper) works pretty well, especially for compacting large arrays. It uses less memory to do it and is slightly faster than my previous method (5-10%). I put in a few tweaks to the poster's algorithm: 1) eliminating the final warp shuffle reduction in phase 1, can simply sum the elements as they are calculated, 2) giving the function the ability to work over more than just arrays sized as a multiple of 1024 + adding grid-strided loops, and 3) allowing each thread to load their registers simultaneously in phase 3 instead of one at a time. I also use CUB instead of Thrust for Inclusive sum for faster scans. There may be more tweaks I can make, but for now this is good.
//kernel phase 1
int myID = blockIdx.x*blockDim.x + threadIdx.x;
//padded_length is nearest multiple of 1024 > true_length
for(int id = myID; id < (padded_length >> 5); id+= blockDim.x*gridDim.x){
int lnID = threadIdx.x % warp_size;
int warpID = id >> 5;
unsigned int mask;
unsigned int cnt=0;//;//
for(int j = 0; j < 32; j++){
int index = (warpID<<10)+(j<<5)+lnID;
bool pred;
if(index > true_length) pred = false;
else pred = predicate(input[index]);
mask = __ballot(pred);
if(lnID == 0) {
flag[(warpID<<5)+j] = mask;
cnt += __popc(mask);
}
}
if(lnID == 0) counter[warpID] = cnt; //store sum
}
//kernel phase 2 -> CUB Inclusive sum transforms counter array to scan_Index array
//kernel phase 3
int myID = blockIdx.x*blockDim.x + threadIdx.x;
for(int id = myID; id < (padded_length >> 5); id+= blockDim.x*gridDim.x){
int lnID = threadIdx.x % warp_size;
int warpID = id >> 5;
unsigned int predmask;
unsigned int cnt;
predmask = flag[(warpID<<5)+lnID];
cnt = __popc(predmask);
//parallel prefix sum
#pragma unroll
for(int offset = 1; offset < 32; offset<<=1){
unsigned int n = __shfl_up(cnt, offset);
if(lnID >= offset) cnt += n;
}
unsigned int global_index = 0;
if(warpID > 0) global_index = scan_Index[warpID - 1];
for(int i = 0; i < 32; i++){
unsigned int mask = __shfl(predmask, i); //broadcast from thread i
unsigned int sub_group_index = 0;
if(i > 0) sub_group_index = __shfl(cnt, i-1);
if(mask & (1 << lnID)){
compacted_array[global_index + sub_group_index + __popc(mask & ((1 << lnID) - 1))] = input[(warpID<<10)+(i<<5)+lnID];
}
}
}
}
EDIT: There is a newer article by a subset of the poster authors where they examine a faster variation of compact than what is written above. However, their new version is not order preserving, so not useful for myself and I haven't implemented it to test it out. That said, if your project doesn't rely on object order, their newer compact version can probably speed up your algorithm.
I am currently writing a code, that calculates a integral Histogram on the GPU using the Nvidia thrust library.
Therefore I allocate a continuous Block of device memory which I update with a custom functor all the time.
The problem is, that the write to the device memory is veeery slow, but the reads are actually ok.
The basic setup is the following:
struct HistogramCreation
{
HistogramCreation(
...
// pointer to memory
...
){}
/// The actual summation operator
__device__ void operator()(int index){
.. do the calculations ..
for(int j=0;j<30;j++){
(1) *_memoryPointer = values (also using reads to such locations) ;
}
}
}
void foo(){
cudaMalloc(_pointer,size);
HistogramCreation initialCreation( ... _pointer ...);
thrust::for_each(
thrust::make_counting_iterator(0),
thrust::make_counting_iterator(_imageSize),
initialCreation);
}
if I change the writing in (1) to the following>
unsigned int val = values;
The performance is much better. THis is the only global memory write I have.
Using the memory write I get about 2s for HD Footage.
using the local variable it takes about 50 ms so about a factor of 40 less.
Why is this so slow? how could I improve it?
Just as #OlegTitov said, frequent load/store with global
memory should be avoided as much as possible. When there's a
situation where it's inevitable, then coalesced memory
access can help the execution process not to get too slow;
however in most cases, histogram calculation is pretty tough
to realize the coalesced access.
While most of the above is basically just restating
#OlegTitov's answer, i'd just like to share about an
investigation i did about finding summation with NVIDIA
CUDA. Actually the result is pretty interesting and i hope
it'll be a helpful information for other xcuda developers.
The experiment was basically to run a speed test of finding
summation with various memory access patterns: using global
memory (1 thread), L2 cache (atomic ops - 128 threads), and
L1 cache (shared mem - 128 threads)
This experiment used:
Kepler GTX 680,
1546 cores # 1.06GHz
GDDR5 256-bit # 3GHz
Here are the kernels:
__global__
void glob(float *h) {
float* hist = h;
uint sd = SEEDRND;
uint random;
for (int i = 0; i < NUMLOOP; i++) {
if (i%NTHREADS==0) random = rnd(sd);
int rind = random % NBIN;
float randval = (float)(random % 10)*1.0f ;
hist[rind] += randval;
}
}
__global__
void atom(float *h) {
float* hist = h;
uint sd = SEEDRND;
for (int i = threadIdx.x; i < NUMLOOP; i+=NTHREADS) {
uint random = rnd(sd);
int rind = random % NBIN;
float randval = (float)(random % 10)*1.0f ;
atomicAdd(&hist[rind], randval);
}
}
__global__
void shm(float *h) {
int lid = threadIdx.x;
uint sd = SEEDRND;
__shared__ float shm[NTHREADS][NBIN];
for (int i = 0; i < NBIN; i++) shm[lid][i] = h[i];
for (int i = lid; i < NUMLOOP; i+=NTHREADS) {
uint random = rnd(sd);
int rind = random % NBIN;
float randval = (float)(random % 10)*1.0f ;
shm[lid][rind] += randval;
}
/* reduction here */
for (int i = 0; i < NBIN; i++) {
__syncthreads();
if (threadIdx.x < 64) {
shm[threadIdx.x][i] += shm[threadIdx.x+64][i];
}
__syncthreads();
if (threadIdx.x < 32) {
shm[threadIdx.x][i] += shm[threadIdx.x+32][i];
}
__syncthreads();
if (threadIdx.x < 16) {
shm[threadIdx.x][i] += shm[threadIdx.x+16][i];
}
__syncthreads();
if (threadIdx.x < 8) {
shm[threadIdx.x][i] += shm[threadIdx.x+8][i];
}
__syncthreads();
if (threadIdx.x < 4) {
shm[threadIdx.x][i] += shm[threadIdx.x+4][i];
}
__syncthreads();
if (threadIdx.x < 2) {
shm[threadIdx.x][i] += shm[threadIdx.x+2][i];
}
__syncthreads();
if (threadIdx.x == 0) {
shm[0][i] += shm[1][i];
}
}
for (int i = 0; i < NBIN; i++) h[i] = shm[0][i];
}
OUTPUT
atom: 102656.00 shm: 102656.00 glob: 102656.00
atom: 122240.00 shm: 122240.00 glob: 122240.00
... blah blah blah ...
One Thread: 126.3919 msec
Atomic: 7.5459 msec
Sh_mem: 2.2207 msec
The ratio between these kernels is 57:17:1. Many things can
be analyzed here, and it truly does not mean that using
L1 or L2 memory spaces will always give you more than 10
times speedup of the whole program.
And here's the main and other funcs:
#include <iostream>
#include <cstdlib>
#include <cstdio>
using namespace std;
#define NUMLOOP 1000000
#define NBIN 36
#define SEEDRND 1
#define NTHREADS 128
#define NBLOCKS 1
__device__ uint rnd(uint & seed) {
#if LONG_MAX > (16807*2147483647)
int const a = 16807;
int const m = 2147483647;
seed = (long(seed * a))%m;
return seed;
#else
double const a = 16807;
double const m = 2147483647;
double temp = seed * a;
seed = (int) (temp - m * floor(temp/m));
return seed;
#endif
}
... the above kernels ...
int main()
{
float *h_hist, *h_hist2, *h_hist3, *d_hist, *d_hist2,
*d_hist3;
h_hist = (float*)malloc(NBIN * sizeof(float));
h_hist2 = (float*)malloc(NBIN * sizeof(float));
h_hist3 = (float*)malloc(NBIN * sizeof(float));
cudaMalloc((void**)&d_hist, NBIN * sizeof(float));
cudaMalloc((void**)&d_hist2, NBIN * sizeof(float));
cudaMalloc((void**)&d_hist3, NBIN * sizeof(float));
for (int i = 0; i < NBIN; i++) h_hist[i] = 0.0f;
cudaMemcpy(d_hist, h_hist, NBIN * sizeof(float),
cudaMemcpyHostToDevice);
cudaMemcpy(d_hist2, h_hist, NBIN * sizeof(float),
cudaMemcpyHostToDevice);
cudaMemcpy(d_hist3, h_hist, NBIN * sizeof(float),
cudaMemcpyHostToDevice);
cudaEvent_t start, end;
float elapsed = 0, elapsed2 = 0, elapsed3;
cudaEventCreate(&start);
cudaEventCreate(&end);
cudaEventRecord(start, 0);
atom<<<NBLOCKS, NTHREADS>>>(d_hist);
cudaThreadSynchronize();
cudaEventRecord(end, 0);
cudaEventSynchronize(start);
cudaEventSynchronize(end);
cudaEventElapsedTime(&elapsed, start, end);
cudaEventRecord(start, 0);
shm<<<NBLOCKS, NTHREADS>>>(d_hist2);
cudaThreadSynchronize();
cudaEventRecord(end, 0);
cudaEventSynchronize(start);
cudaEventSynchronize(end);
cudaEventElapsedTime(&elapsed2, start, end);
cudaEventRecord(start, 0);
glob<<<1, 1>>>(d_hist3);
cudaThreadSynchronize();
cudaEventRecord(end, 0);
cudaEventSynchronize(start);
cudaEventSynchronize(end);
cudaEventElapsedTime(&elapsed3, start, end);
cudaMemcpy(h_hist, d_hist, NBIN * sizeof(float),
cudaMemcpyDeviceToHost);
cudaMemcpy(h_hist2, d_hist2, NBIN * sizeof(float),
cudaMemcpyDeviceToHost);
cudaMemcpy(h_hist3, d_hist3, NBIN * sizeof(float),
cudaMemcpyDeviceToHost);
/* print output */
for (int i = 0; i < NBIN; i++) {
printf("atom: %10.2f shm: %10.2f glob:
%10.2f¥n",h_hist[i],h_hist2[i],h_hist3[i]);
}
printf("%12s: %8.4f msec¥n", "One Thread", elapsed3);
printf("%12s: %8.4f msec¥n", "Atomic", elapsed);
printf("%12s: %8.4f msec¥n", "Sh_mem", elapsed2);
return 0;
}
When writing GPU code you should avoid reading and writing to/from global memory. Global memory is very slow on GPU. That's the hardware feature. The only thing you can do is to make neighboring treads read/write in neighboring adresses in global memory. This will cause coalescing and speed up the process. But in general read your data once, process it and write it out once.
Note that NVCC might optimize out a lot of your code after you make the modification - it detects that no write to global memory is made and just removes the "unneeded" code. So this speedup may not be coming out of the global writer per ce.
I would recommend using profiler on your actual code (the one with global write) to see if there's anything like unaligned access or other perf problem.
OK, so lets say I have an ( N x N ) matrix that I would like to process. This matrix is quite large for my computer, and if I try to send it to the device all at once I get a 'out of memory error.'
So is there a way to send sections of the matrix to the device? One way I can see to do it is copy portions of the matrix on the host, and then send these manageable copied portions from the host to the device, and then put them back together at the end.
Here is something I have tried, but the cudaMemcpy in the for loop returns error code 11, 'invalid argument.'
int h_N = 10000;
size_t h_size_m = h_N*sizeof(float);
h_A = (float*)malloc(h_size_m*h_size_m);
int d_N = 2500;
size_t d_size_m = d_N*sizeof(float);
InitializeMatrices(h_N);
int i;
int iterations = (h_N*h_N)/(d_N*d_N);
for( i = 0; i < iterations; i++ )
{
float* h_array_ref = h_A+(i*d_N*d_N);
cudasafe( cudaMemcpy(d_A, h_array_ref, d_size_m*d_size_m, cudaMemcpyHostToDevice), "cudaMemcpy");
cudasafe( cudaFree(d_A), "cudaFree(d_A)" );
}
What I'm trying to accomplish with the above code is this: instead of send the entire matrix to the device, I simply send a pointer to a place within that matrix and reserve enough space on the device to do the work, and then with the next iteration of the loop move the pointer forward within the matrix, etc. etc.
Not only can you do this (assuming your problem is easily decomposed this way into sub-arrays), it can be a very useful thing to do for performance; once you get the basic approach you've described working, you can start using asynchronous memory copies and double-buffering to overlap some of the memory transfer time with the time spent computing what is already on-card.
But first one gets the simple thing working. Below is a 1d example (multiplying a vector by a scalar and adding another scalar) but using a linearized 2d array would be the same; the key part is
CHK_CUDA( cudaMalloc(&xd, batchsize*sizeof(float)) );
CHK_CUDA( cudaMalloc(&yd, batchsize*sizeof(float)) );
tick(&gputimer);
int nbatches = 0;
for (int nstart=0; nstart < n; nstart+=batchsize) {
int size=batchsize;
if ((nstart + batchsize) > n) size = n - nstart;
CHK_CUDA( cudaMemcpy(xd, &(x[nstart]), size*sizeof(float), cudaMemcpyHostToDevice) );
blocksize = (size+nblocks-1)/nblocks;
cuda_saxpb<<<nblocks, blocksize>>>(xd, a, b, yd, size);
CHK_CUDA( cudaMemcpy(&(ycuda[nstart]), yd, size*sizeof(float), cudaMemcpyDeviceToHost) );
nbatches++;
}
gputime = tock(&gputimer);
CHK_CUDA( cudaFree(xd) );
CHK_CUDA( cudaFree(yd) );
You allocate the buffers at the start, and then loop through until you're done, each time doing the copy, starting the kernel, and then copying back. You free at the end.
The full code is
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
#include <cuda.h>
#include <sys/time.h>
#include <math.h>
#define CHK_CUDA(e) {if (e != cudaSuccess) {fprintf(stderr,"Error: %s\n", cudaGetErrorString(e)); exit(-1);}}
__global__ void cuda_saxpb(const float *xd, const float a, const float b,
float *yd, const int n) {
int i = threadIdx.x + blockIdx.x*blockDim.x;
if (i<n) {
yd[i] = a*xd[i]+b;
}
return;
}
void cpu_saxpb(const float *x, float a, float b, float *y, int n) {
int i;
for (i=0;i<n;i++) {
y[i] = a*x[i]+b;
}
return;
}
int get_options(int argc, char **argv, int *n, int *s, int *nb, float *a, float *b);
void tick(struct timeval *timer);
double tock(struct timeval *timer);
int main(int argc, char **argv) {
int n=1000;
int nblocks=10;
int batchsize=100;
float a = 5.;
float b = -1.;
int err;
float *x, *y, *ycuda;
float *xd, *yd;
double abserr;
int blocksize;
int i;
struct timeval cputimer;
struct timeval gputimer;
double cputime, gputime;
err = get_options(argc, argv, &n, &batchsize, &nblocks, &a, &b);
if (batchsize > n) {
fprintf(stderr, "Resetting batchsize to size of vector, %d\n", n);
batchsize = n;
}
if (err) return 0;
x = (float *)malloc(n*sizeof(float));
if (!x) return 1;
y = (float *)malloc(n*sizeof(float));
if (!y) {free(x); return 1;}
ycuda = (float *)malloc(n*sizeof(float));
if (!ycuda) {free(y); free(x); return 1;}
/* run CPU code */
tick(&cputimer);
cpu_saxpb(x, a, b, y, n);
cputime = tock(&cputimer);
/* run GPU code */
/* only have to allocate once */
CHK_CUDA( cudaMalloc(&xd, batchsize*sizeof(float)) );
CHK_CUDA( cudaMalloc(&yd, batchsize*sizeof(float)) );
tick(&gputimer);
int nbatches = 0;
for (int nstart=0; nstart < n; nstart+=batchsize) {
int size=batchsize;
if ((nstart + batchsize) > n) size = n - nstart;
CHK_CUDA( cudaMemcpy(xd, &(x[nstart]), size*sizeof(float), cudaMemcpyHostToDevice) );
blocksize = (size+nblocks-1)/nblocks;
cuda_saxpb<<<nblocks, blocksize>>>(xd, a, b, yd, size);
CHK_CUDA( cudaMemcpy(&(ycuda[nstart]), yd, size*sizeof(float), cudaMemcpyDeviceToHost) );
nbatches++;
}
gputime = tock(&gputimer);
CHK_CUDA( cudaFree(xd) );
CHK_CUDA( cudaFree(yd) );
abserr = 0.;
for (i=0;i<n;i++) {
abserr += fabs(ycuda[i] - y[i]);
}
printf("Y = a*X + b, problemsize = %d\n", n);
printf("CPU time = %lg millisec.\n", cputime*1000.);
printf("GPU time = %lg millisec (done with %d batches of %d).\n",
gputime*1000., nbatches, batchsize);
printf("CUDA and CPU results differ by %lf\n", abserr);
free(x);
free(y);
free(ycuda);
return 0;
}
int get_options(int argc, char **argv, int *n, int *s, int *nb, float *a, float *b) {
const struct option long_options[] = {
{"nvals" , required_argument, 0, 'n'},
{"nblocks" , required_argument, 0, 'B'},
{"batchsize" , required_argument, 0, 's'},
{"a", required_argument, 0, 'a'},
{"b", required_argument, 0, 'b'},
{"help", no_argument, 0, 'h'},
{0, 0, 0, 0}};
char c;
int option_index;
int tempint;
while (1) {
c = getopt_long(argc, argv, "n:B:a:b:s:h", long_options, &option_index);
if (c == -1) break;
switch(c) {
case 'n': tempint = atoi(optarg);
if (tempint < 1 || tempint > 500000) {
fprintf(stderr,"%s: Cannot use number of points %s;\n Using %d\n", argv[0], optarg, *n);
} else {
*n = tempint;
}
break;
case 's': tempint = atoi(optarg);
if (tempint < 1 || tempint > 50000) {
fprintf(stderr,"%s: Cannot use number of points %s;\n Using %d\n", argv[0], optarg, *s);
} else {
*s = tempint;
}
break;
case 'B': tempint = atoi(optarg);
if (tempint < 1 || tempint > 1000 || tempint > *n) {
fprintf(stderr,"%s: Cannot use number of blocks %s;\n Using %d\n", argv[0], optarg, *nb);
} else {
*nb = tempint;
}
break;
case 'a': *a = atof(optarg);
break;
case 'b': *b = atof(optarg);
break;
case 'h':
puts("Calculates y[i] = a*x[i] + b on the GPU.");
puts("Options: ");
puts(" --nvals=N (-n N): Set the number of values in y,x.");
puts(" --batchsize=N (-s N): Set the number of values to transfer at a time.");
puts(" --nblocks=N (-B N): Set the number of blocks used.");
puts(" --a=X (-a X): Set the parameter a.");
puts(" --b=X (-b X): Set the parameter b.");
puts(" --niters=N (-I X): Set number of iterations to calculate.");
puts("");
return +1;
}
}
return 0;
}
void tick(struct timeval *timer) {
gettimeofday(timer, NULL);
}
double tock(struct timeval *timer) {
struct timeval now;
gettimeofday(&now, NULL);
return (now.tv_usec-timer->tv_usec)/1.0e6 + (now.tv_sec - timer->tv_sec);
}
Running this one gets:
$ ./batched-saxpb --nvals=10240 --batchsize=10240 --nblocks=20
Y = a*X + b, problemsize = 10240
CPU time = 0.072 millisec.
GPU time = 0.117 millisec (done with 1 batches of 10240).
CUDA and CPU results differ by 0.000000
$ ./batched-saxpb --nvals=10240 --batchsize=5120 --nblocks=20
Y = a*X + b, problemsize = 10240
CPU time = 0.066 millisec.
GPU time = 0.133 millisec (done with 2 batches of 5120).
CUDA and CPU results differ by 0.000000
$ ./batched-saxpb --nvals=10240 --batchsize=2560 --nblocks=20
Y = a*X + b, problemsize = 10240
CPU time = 0.067 millisec.
GPU time = 0.167 millisec (done with 4 batches of 2560).
CUDA and CPU results differ by 0.000000
The GPU time goes up in this case (we're doing more memory copies) but the answers stay the same.
Edited: The original version of this code had an option for running multiple iterations of the kernel for timing purposes, but that's unnecessarily confusing in this context so it's removed.
I'm trying to understand the impact of strict aliasing on performance in C99. My goal is to optimize a vector dot product, which takes up a large amount of time in my program (profiled it!). I thought that aliasing could be the problem, but the following code doesn't show any substantial difference between the standard approach and the strict aliasing version, even with vectors of size 100 million. I've also tried to use local variables to avoid aliasing, with similar results.
What's happening?
I'm using gcc-4.7 on OSX 10.7.4. Results are in microseconds.
$ /usr/local/bin/gcc-4.7 -fstrict-aliasing -Wall -std=c99 -O3 -o restrict restrict.c
$ ./restrict
sum: 100000000 69542
sum2: 100000000 70432
sum3: 100000000 70372
sum4: 100000000 69891
$ /usr/local/bin/gcc-4.7 -Wall -std=c99 -O0 -fno-strict-aliasing -o restrict restrict.c
$ ./restrict
sum: 100000000 258487
sum2: 100000000 261349
sum3: 100000000 258829
sum4: 100000000 258129
restrict.c (note this code will need several hundred MB RAM):
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <sys/time.h>
#include <unistd.h>
/* original */
long sum(int *x, int *y, int n)
{
long i, s = 0;
for(i = 0 ; i < n ; i++)
s += x[i] * y[i];
return s;
}
/* restrict */
long sum2(int *restrict x, int *restrict y, int n)
{
long i, s = 0;
for(i = 0 ; i < n ; i++)
s += x[i] * y[i];
return s;
}
/* local restrict */
long sum3(int *x, int *y, int n)
{
int *restrict xr = x;
int *restrict yr = y;
long i, s = 0;
for(i = 0 ; i < n ; i++)
s += xr[i] * yr[i];
return s;
}
/* use local variables */
long sum4(int *x, int *y, int n)
{
int xr, yr;
long i, s = 0;
for(i = 0 ; i < n ; i++)
{
xr = x[i];
yr = y[i];
s += xr * yr;
}
return s;
}
int main(void)
{
struct timeval tp1, tp2;
struct timezone tzp;
long i, n = 1e8L, s;
int *x = malloc(sizeof(int) * n);
int *y = malloc(sizeof(int) * n);
long elapsed1;
for(i = 0 ; i < n ; i++)
x[i] = y[i] = 1;
gettimeofday(&tp1, &tzp);
s = sum(x, y, n);
gettimeofday(&tp2, &tzp);
elapsed1 = (tp2.tv_sec - tp1.tv_sec) * 1e6
+ (tp2.tv_usec - tp1.tv_usec);
printf("sum:\t%ld\t%ld\n", s, elapsed1);
gettimeofday(&tp1, &tzp);
s = sum2(x, y, n);
gettimeofday(&tp2, &tzp);
elapsed1 = (tp2.tv_sec - tp1.tv_sec) * 1e6
+ (tp2.tv_usec - tp1.tv_usec);
printf("sum2:\t%ld\t%ld\n", s, elapsed1);
gettimeofday(&tp1, &tzp);
s = sum3(x, y, n);
gettimeofday(&tp2, &tzp);
elapsed1 = (tp2.tv_sec - tp1.tv_sec) * 1e6
+ (tp2.tv_usec - tp1.tv_usec);
printf("sum3:\t%ld\t%ld\n", s, elapsed1);
gettimeofday(&tp1, &tzp);
s = sum3(x, y, n);
gettimeofday(&tp2, &tzp);
elapsed1 = (tp2.tv_sec - tp1.tv_sec) * 1e6
+ (tp2.tv_usec - tp1.tv_usec);
printf("sum4:\t%ld\t%ld\n", s, elapsed1);
return EXIT_SUCCESS;
}
Off the cuff:
with no strict aliasing rules, the compiler might simply generate optimized code that does subtly different things than intended.
It is not a given that disabling strict aliasing rules leads to faster code.
If it does, it's also not a given that the optimized code actually show different results. This depends a lot on the actual data access patterns, and often even the processor/cache architecture.
Regarding your example code, I'd say that aliasing is irrelevant (for emitted code, at least) since there is never any write access to the array elements inside the sumXXX functions.
(You might get slightly better performance (or opposite) if you pass the same vector twice. There might be a boon from hot cache and smaller cache footprint. There may be a penalty from redundant Loads putting the prefetch predictor off-track. As always: use a profiler)