Zero padding on the fly with cuFFT - cuda

I have a float array and want to FFT from this with an amount of data and padding by zero padding to 2^N. I also want to overlap the data by a selectable factor.
So far I have a cuda kernel with which I create another array in which I store the overlapped and padded data. Afterwards a cufftPlanMany is executed.
By the two factors, the amount of data becomes very large and it is in principle only copies of the original data and zeros with which I waste my entire memory bandwidth.
I could not find anything if cuFFT supports zero padding or if I have a possibility to create custom scripts.
(Nvidia Quadro P5000, C++14, Kubuntu)
Update
I have written a callback function which is called when loading the data into the FFT. Unfortunately this is still a little bit slower than my previous solution with a kernel which prepares the data in another array and then calls the FFT.
I need an average of 2.4ms for the example with the given values.
My hope was that if I process the data on the fly, my memory bandwidth will not limit me anymore. Unfortunately this does not look like that at the moment.
Does anyone have an idea how I can speed this up even more?
// Don't forget to include cufft_static(not cufft), culibos and set flag -dc
#include <stdio.h>
#include <cstdint>
#include <unistd.h>
#include <cuda_runtime.h>
#include <cufft.h>
#include <cufftXt.h>
#include <math.h>
typedef struct fft_CB_LD_callerInfo{
uint16_t rgLen;
uint16_t rgDataLen;
uint16_t overlapFactor;
};
static __device__ cufftReal myOwnCallback(void *dataIn,
size_t offset,
void *callerInfo,
void *sharedPtr) {
const fft_CB_LD_callerInfo *fftInfo = (fft_CB_LD_callerInfo*)callerInfo;
int idx_rg = offset/fftInfo->rgLen;
int idx_realRg = idx_rg/fftInfo->overlapFactor;
int idx_posInRg = offset-(size_t)idx_rg*fftInfo->rgLen;
if(idx_posInRg < fftInfo->rgDataLen){
const size_t idx_data = idx_posInRg
+ idx_realRg*fftInfo->rgDataLen
+ idx_rg - (idx_realRg*fftInfo->overlapFactor)*fftInfo->rgDataLen/fftInfo->overlapFactor;
return ((cufftReal*)dataIn)[idx_data];
}
else{
return 0.0f;
}
}
__device__ cufftCallbackLoadR myOwnCallbackPtr = myOwnCallback;
int main(){
// Data
float *dataHost;
float *data;
cufftComplex *spectrum;
cufftComplex *spectrumHost;
unsigned int rgDataLen = 400;
unsigned int rgLen = 2048;
unsigned int overlap = 8;
int peakPosHost[] = {0};
int *peakPos;
unsigned int rgCountClean = 52*16*4;
unsigned int rgCount = rgCountClean*overlap-(overlap-1);
int peakCountHost = 1;
int *peakCount;
// for FFT
cudaStream_t stream;
cufftHandle plan;
cufftResult result;
int fftRank = 1; // --- 1D FFTs
int fftIRide = 1, fftORide = 1; // --- Distance between two successive input/output elements
int fftInembed[] = { 0 }; // --- Input size with pitch (ignored for 1D transforms)
int fftOnembed[] = { 0 }; // --- Output size with pitch (ignored for 1D transforms)
int fftEachLen[] = { (int)rgLen }; // --- Size of the Fourier transform
int fftIDist = rgLen;
int fftODist = rgLen/2+1; // --- Distance between batches
// for Custom callback
cufftCallbackLoadR hostCopyOfCallbackPtr;
size_t worksize;
fft_CB_LD_callerInfo *fftInfo;
fft_CB_LD_callerInfo *fftInfoHost;
// Allocate host memory
dataHost = new float[rgDataLen*rgCountClean*peakCountHost];
spectrumHost = new cufftComplex[fftODist*rgCount];
fftInfoHost = new fft_CB_LD_callerInfo;
// create array with example data
for(int k=0; k<rgDataLen;k++){
for(int i=0; i<rgCountClean; i++){
dataHost[i*rgDataLen + k] = sin((2+i*4)*M_PI*k/rgDataLen);
}
}
fftInfoHost->overlapFactor = overlap;
fftInfoHost->rgDataLen = rgDataLen;
fftInfoHost->rgLen = rgLen;
// allocate device memory
cudaMalloc((void **)&data, sizeof(float) * rgDataLen*rgCountClean*peakCountHost);
cudaMalloc((void **)&peakPos, sizeof(int) * peakCountHost);
cudaMalloc((void **)&peakCount, sizeof(int));
cudaMalloc((void **)&spectrum, sizeof(cufftComplex)*fftODist*rgCount);
cudaMalloc((void **)&fftInfo, sizeof(fft_CB_LD_callerInfo));
// copy date from host to device
cudaMemcpy(data, dataHost, sizeof(float)*rgDataLen*rgCountClean*peakCountHost, cudaMemcpyHostToDevice);
cudaMemcpy(peakPos, peakPosHost, sizeof(int)*peakCountHost, cudaMemcpyHostToDevice);
cudaMemcpy(peakCount, &peakCountHost, sizeof(peakCountHost), cudaMemcpyHostToDevice);
cudaMemcpy(fftInfo, fftInfoHost, sizeof(fft_CB_LD_callerInfo), cudaMemcpyHostToDevice);
// get device pointer to custom callback function
cudaError_t error = cudaMemcpyFromSymbol(&hostCopyOfCallbackPtr, myOwnCallbackPtr, sizeof(hostCopyOfCallbackPtr));
if(error != 0) printf("cudaMemcpyFromSymbol faild with %d!\n", (int)error);
// Create a plan of FFTs to fast execute there later
cufftCreate(&plan);
result = cufftMakePlanMany(plan, fftRank, fftEachLen, fftInembed, fftIRide, fftIDist, fftOnembed, fftORide, fftODist, CUFFT_R2C, rgCount, &worksize);
if(result != CUFFT_SUCCESS) printf("cufftMakePlanMany failed with %d!\n", (int)result);
result = cufftXtSetCallback(plan, (void**)&hostCopyOfCallbackPtr, CUFFT_CB_LD_REAL, (void**)&fftInfo);
if(result != CUFFT_SUCCESS) printf("cufftXtSetCallback failed with %d!\n", (int)result);
// ----- Begin test area ---------------------------------------------------
if(cufftExecR2C(plan, data, spectrum) != CUFFT_SUCCESS)
printf("cufftExecR2C is failed!\n");
// ----- End test area ---------------------------------------------------
return 0;
}

Related

creating 16-bit input to cufftXtMakePlanMany and workSize for 1 GPU

I need to compute FFT on unsigned int 8bit data. Previously, I was using cufftPlanMany and my input was cufftReal and the output was cufftComplex, and I was using casting before and after FFT to convert from unsigned 8bit to cufftReal and then from cufftComplex to signed 8bit.
It came to my attention that cuFFT has a nice option to run FFT for half-precision data which I hope improves the running time. According to the documentation, it currently doesn't support all of cudaDataType (that would be wonderful if it can in the future), but at least I can run it with 16bit float (half-precision) with the following signature:
cufftResult
cufftXtMakePlanMany(cufftHandle plan, int rank, long long int *n, long long int *inembed,
long long int istride, long long int idist, cudaDataType inputtype,
long long int *onembed, long long int ostride, long long int odist,
cudaDataType outputtype, long long int batch, size_t *workSize,
cudaDataType executiontype);
with data types for input, output and execution respectively as: CUDA_R_16F, CUDA_C_16F and CUDA_C_16F. Tha twould be ideal that I can feed this cuFFT with my U8 data, is there any way for doing so? Otherwise, if the first casting from U8 to cufftReal is necessary how can I convert my data from cufftReal to CUDA_R_16F and then from CUDA_C_16F ? Is cuda smart enough to cast the input from float to half-precision data, because cufftExecR2C ultimaltey would be the same and there is no other function to be called for the half-precision?
The other question is about workSize which is designed for multiple GPU cases. Any idea how this size has to be calculated? (I have just 1 GPU). Am I responsible for managing that buffer?
TL;DR: I can see two possible approaches here, one using a half-precision transform and one using a single-precision transform (perhaps with CUFFT callbacks). The reasons to choose one or the other may depend on a number of factors such as size of your transform, control of scope of input data, the GPU you are running on, and other factors.
I'm not going to try to address the processing of the output data that you indicate here:
then from cufftComplex to signed 8bit.
since I don't know how to do that without more information. However the processing of the input data in each case should be illustrative for how you could process the output data.
Using half-precision transforms
A few things to note here are that you cannot (currently) use callbacks with half-precision transforms, and half-precision transforms can be more sensitive to input data characteristics (e.g. DC offset, transform size, etc.) than single or double-precision transforms. Also, half-precision transforms for the most part require a pascal or newer GPU (ignoring Jetson family).
Because half-precision transforms don't support callbacks, we'll use "ordinary" host code to process the input data; you could also do this processing on the device prior to the transform, the provided code outlines both possibilities. My "preprocessing" here is mostly just designed to prevent the 16-bit transform from overflowing. If you play around with this code you'll quickly see what an overflow looks like (inf and/or nan in the output).
$ cat t1961.cu
#include <cufft.h>
#include <stdio.h>
#include <stdlib.h>
#include <cufftXt.h>
#include <cuda_fp16.h>
#include <assert.h>
#include <iostream>
typedef half2 ctype;
typedef half rtype;
typedef unsigned char dtype;
long long sig_size = 1<<18;
const int amplitude = 127;
const float ramplitude = 1/(float)(4*amplitude);
__host__ __device__ half convert(int val){
return __float2half_rn((val - amplitude)*ramplitude);
}
__global__ void dev_convert(rtype *out, dtype *in, int sz){
int idx = threadIdx.x+blockDim.x*blockIdx.x;
if (idx < sz)
out[idx] = convert(in[idx]);
}
int main(){
//put 4x sine waves into a U8 array
dtype *my_data = (dtype *)malloc(sig_size*sizeof(dtype));
for (int i = 0; i < sig_size; i++) my_data[i] = amplitude*(sin((i*8*3.141592654f)/sig_size)+1.0);
rtype *d_idata;
ctype *d_odata;
cudaMalloc(&d_idata, sizeof(rtype)*sig_size);
#ifdef USE_HOST
rtype *h_idata = (rtype *)malloc(sig_size*sizeof(rtype));
// convert to 16 bit float non-offset suitable for cufft
for (int i = 0; i < sig_size; i++) h_idata[i] = convert(my_data[i]);
cudaMemcpy(d_idata, h_idata, sig_size*sizeof(rtype), cudaMemcpyHostToDevice);
#else
const int bs = 256;
dtype *d_mydata;
cudaMalloc(&d_mydata, sig_size*sizeof(dtype));
cudaMemcpy(d_mydata, my_data, sig_size*sizeof(dtype), cudaMemcpyHostToDevice);
dev_convert<<<(sig_size+bs-1)/bs, bs>>>(d_idata, d_mydata, sig_size);
#endif
cudaMalloc(&d_odata, sizeof(ctype)*(sig_size/2+1));
cufftHandle plan;
cufftResult r;
r = cufftCreate(&plan);
assert(r == CUFFT_SUCCESS);
size_t ws = 0;
r = cufftXtMakePlanMany(plan, 1, &sig_size, NULL, 1, 1, CUDA_R_16F, NULL, 1, 1, CUDA_C_16F, 1, &ws, CUDA_C_16F);
assert(r == CUFFT_SUCCESS);
r = cufftXtExec(plan, d_idata, d_odata, CUFFT_FORWARD); // warm-up
assert(r == CUFFT_SUCCESS);
cudaEvent_t start, stop;
cudaEventCreate(&start); cudaEventCreate(&stop);
cudaEventRecord(start);
r = cufftXtExec(plan, d_idata, d_odata, CUFFT_FORWARD);
assert(r == CUFFT_SUCCESS);
cudaEventRecord(stop);
cudaEventSynchronize(stop);
float et;
cudaEventElapsedTime(&et, start, stop);
printf("forward FFT time for %ld samples: %fms\n", sig_size, et);
ctype *h_odata = (ctype *)malloc((sig_size/2+1)*sizeof(ctype));
cudaMemcpy(h_odata, d_odata, (sig_size/2+1)*sizeof(ctype), cudaMemcpyDeviceToHost);
for (int i = 0; i < 8; i++)
std::cout << __half2float(h_odata[i].x) << " + " << __half2float(h_odata[i].y) << "i" << std::endl;
return 0;
}
$ nvcc -o t1961 t1961.cu -lcufft
$ ./t1961
forward FFT time for 262144 samples: 0.027520ms
-258 + 0i
0.00349998 + 0.00127506i
-0.000146866 + -0.000833511i
0.00140095 + -0.00501251i
-1.57031 + -32752i
-0.00198174 + 0.00856018i
0.00474548 + 0.00359917i
-0.00226784 + 0.00987244i
$
Using a single precision transform with a load callback
This in my view has a few benefits. It is not as subject to the overflow phenomenon as the half precision transforms are, and the (load) callback routine allows us to still operate on U8 input data.
$ cat t1962.cu
#include <cufft.h>
#include <stdio.h>
#include <stdlib.h>
#include <cufftXt.h>
#include <cuda_fp16.h>
#include <assert.h>
#include <iostream>
typedef cufftComplex ctype;
typedef cufftReal rtype;
typedef unsigned char dtype;
long long sig_size = 1<<18;
const int amplitude = 127;
const cufftReal ramplitude = 1/(float)(4*amplitude);
__device__ rtype convert(int val){
return (val - amplitude)*ramplitude;
}
__device__ rtype myOwnCallback(void *dataIn,
size_t offset,
void *callerInfo,
void *sharedPtr) {
rtype ret;
ret = convert(((dtype *)dataIn)[offset]);
return ret;
}
__device__ cufftCallbackLoadR myOwnCallbackPtr = myOwnCallback;
int main(){
cufftCallbackLoadR hostCopyOfCallbackPtr;
cudaMemcpyFromSymbol(&hostCopyOfCallbackPtr,
myOwnCallbackPtr,
sizeof(hostCopyOfCallbackPtr));
//put 4x sine waves into a U8 array
dtype *my_data = (dtype *)malloc(sig_size*sizeof(dtype));
for (int i = 0; i < sig_size; i++) my_data[i] = amplitude*(sin((i*8*3.141592654f)/sig_size)+1.0);
ctype *d_odata;
dtype *d_mydata;
cudaMalloc(&d_mydata, sig_size*sizeof(dtype));
cudaMemcpy(d_mydata, my_data, sig_size*sizeof(dtype), cudaMemcpyHostToDevice);
cudaMalloc(&d_odata, sizeof(ctype)*(sig_size/2+1));
cufftHandle plan;
cufftResult r;
r = cufftCreate(&plan);
assert(r == CUFFT_SUCCESS);
size_t ws = 0;
r = cufftXtMakePlanMany(plan, 1, &sig_size, NULL, 1, 1, CUDA_R_32F, NULL, 1, 1, CUDA_C_32F, 1, &ws, CUDA_C_32F);
assert(r == CUFFT_SUCCESS);
void *rps[] = {(void *)hostCopyOfCallbackPtr};
r = cufftXtSetCallback(plan, rps, CUFFT_CB_LD_REAL, NULL);
assert(r == CUFFT_SUCCESS);
r = cufftXtExec(plan, (cufftReal *)d_mydata, d_odata, CUFFT_FORWARD); // warm-up
assert(r == CUFFT_SUCCESS);
cudaEvent_t start, stop;
cudaEventCreate(&start); cudaEventCreate(&stop);
cudaEventRecord(start);
r = cufftXtExec(plan, (cufftReal *)d_mydata, d_odata, CUFFT_FORWARD);
assert(r == CUFFT_SUCCESS);
cudaEventRecord(stop);
cudaEventSynchronize(stop);
float et;
cudaEventElapsedTime(&et, start, stop);
printf("forward FFT time for %ld samples: %fms\n", sig_size, et);
ctype *h_odata = (ctype *)malloc((sig_size/2+1)*sizeof(ctype));
cudaMemcpy(h_odata, d_odata, (sig_size/2+1)*sizeof(ctype), cudaMemcpyDeviceToHost);
for (int i = 0; i < 8; i++)
std::cout << h_odata[i].x << " + " << h_odata[i].y << "i" << std::endl;
return 0;
}
$ nvcc -o t1962 t1962.cu -rdc=true -lcufft_static -lculibos
$ ./t1962
forward FFT time for 262144 samples: 0.031488ms
-257.969 + 0i
0.00344251 + 0.00137726i
-3.96543e-05 + -0.00106905i
0.0013994 + -0.00490045i
0.0331312 + -32759.4i
-0.00190887 + 0.00865401i
0.00454092 + 0.00368094i
-0.00219025 + 0.00983646i
$
Yes, the results are not numerically identical between the two transform types. It's not reasonable to expect that 16-bit floating point calculations and 32-bit floating point calculations will be identical. In all probability the 32-bit calculations are "more accurate". For this sinewave case, the terms I consider most important are the DC term as well as the magnitude spike at the fundamental. Those are numerically close to each other. The other terms are "in the noise". The timing results are not exactly comparable either, as the 16-bit calculation case omits the cost of the kernel call to convert the data from U8 to F16. You can use a profiler or just refactor the code to get more comparable timing.
workSize can be ignored for the single GPU case when using cufftXtMakePlanMany, otherwise, use the provided routines to determine workSize.

Does bool variable in kernel need to be synchronized

I have a kernel consisting of a for loop that searches through an array for a specific int value. I'm using a grid block of 256 threads to do this. However, when one thread finds the value, I want to let the other threads know to exit. Currently I'm using a boolean flag, but I'm not sure if its working properly. My concern is synchronization.
__device__ bool found;
__global__
void search()
{
for(int i = threadIdx.x; i<1000000; i += stride)
{
if(found == true)
{
break;
}
else if(arr[i] = x)
{
found = true;
break;
}
}
}
int main()
{
bool flag = false;
cudaMemcpyToSymbol(found, &flag, sizeof(bool), 0,cudaMemcpyHostToDevice);
}
As pointed out in comments, you can probably achieve what you want by declaring the global device flag to be volatile, which will inhibit caching, and by using a memory fence function. There really isn't a global synchronization primitive which would do want you want other than the new grid synchronization mechanism introduced in CUDA 9 and new hardware, but that probably isn't necessary in this case. Turning your pseudocode into a toy example:
#include <iostream>
#include <thrust/device_vector.h>
__device__ volatile bool found;
__device__ volatile size_t idx;
template<bool docheck>
__global__
void search(const int* arr, int x, size_t N)
{
size_t i = threadIdx.x + blockIdx.x * blockDim.x;
size_t stride = blockDim.x * gridDim.x;
for(; (i<N) && (!found); i += stride)
{
if(arr[i] == x)
{
if (docheck) found = true;
idx = i;
__threadfence();
break;
}
}
}
int main()
{
const size_t N = 1 << 24;
const size_t findidx = 280270;
const int findval = 0xdeadbeef;
thrust::device_vector<int> data(N,1);
data[findidx] = findval;
bool flag = false;
size_t zero = 0;
{
cudaMemcpyToSymbol(found, &flag, sizeof(bool));
cudaMemcpyToSymbol(idx, &zero, sizeof(size_t));
int blocks, threads;
cudaOccupancyMaxPotentialBlockSize(&blocks, &threads, search<false>);
search<false><<<blocks, threads>>>(thrust::raw_pointer_cast(data.data()), findval, N);
cudaDeviceSynchronize();
size_t result = 0;
cudaMemcpyFromSymbol(&result, idx, sizeof(size_t));
std::cout << "result = " << result << std::endl;
}
{
cudaMemcpyToSymbol(found, &flag, sizeof(bool));
cudaMemcpyToSymbol(idx, &zero, sizeof(size_t));
int blocks, threads;
cudaOccupancyMaxPotentialBlockSize(&blocks, &threads, search<true>);
search<true><<<blocks, threads>>>(thrust::raw_pointer_cast(data.data()), findval, N);
cudaDeviceSynchronize();
size_t result = 0;
cudaMemcpyFromSymbol(&result, idx, sizeof(size_t));
std::cout << "result = " << result << std::endl;
}
return 0;
}
and profiling it gives the following:
$ nvcc -arch=sm_52 -o notify notify.cu
$ nvprof ./notify
==3916== NVPROF is profiling process 3916, command: ./notify
result = 280270
result = 280270
==3916== Profiling application: ./notify
==3916== Profiling result:
Type Time(%) Time Calls Avg Min Max Name
GPU activities: 78.00% 1.6773ms 1 1.6773ms 1.6773ms 1.6773ms void search<bool=0>(int const *, int, unsigned long)
19.93% 428.63us 1 428.63us 428.63us 428.63us void thrust::cuda_cub::core::_kernel_agent<thrust::cuda_cub::__parallel_for::ParallelForAgent<thrust::cuda_cub::__uninitialized_fill::functor<thrust::device_ptr<int>, int>, unsigned long>, thrust::cuda_cub::__uninitialized_fill::functor<thrust::device_ptr<int>, int>, unsigned long>(thrust::device_ptr<int>, int)
1.82% 39.199us 1 39.199us 39.199us 39.199us void search<bool=1>(int const *, int, unsigned long)
As you can see, the version which sets the found flag completes the search in 40 microseconds, whereas the version which does not set the flag takes 1.7 milliseconds. Given that the kernel is run with the maximum number of resident blocks in both cases, we can conclude that the early exit mechanism worked correctly and running blocks detected that the required value had been found.

Atomic Operation failed in CUDA

As the compute ability is 2.1, the atomicAdd and atomicMax operations do not support double precision, then I define both functions based on some answers on stack overflow.
It is strange that the atomicAdd function works well but the atomicMax doesn't work, here is my code.
The test of my code is to generate random number on each block, and then sum the random numbers on each block, we have block sum, I want to test the atomicAdd and atomicMax on the block sum.
#include <iostream>
#include <curand.h>
#include <curand_kernel.h>
#include <stdio.h>
#include <stdlib.h>
#define num_of_blocks 2
#define threads_per_block 2
#define tot_threads 4
__device__ double gsum[num_of_blocks];
__device__ double dev_sum;
__device__ double dev_max;
// set seed for random number generator
__global__ void initcuRand(curandState* globalState, unsigned long seed){
int idx = threadIdx.x + blockIdx.x * blockDim.x;
curand_init(seed, idx, 0, &globalState[idx]);
}
// atomiMax for double
__device__ double atomicMax_d(double* address, double val)
{
unsigned long long int* address_as_i = (unsigned long long int*)address;
unsigned long long int old = *address_as_i, assumed;
do {
assumed = old;
old = ::atomicCAS(address_as_i, assumed, __double_as_longlong(::fmax(val, __longlong_as_double(assumed))));
} while (assumed != old);
return __longlong_as_double(old);
}
// atomicAdd for double
__device__ double atomicAdd_d(double* address, double val)
{
unsigned long long int* address_as_ull = (unsigned long long int*)address;
unsigned long long int old = *address_as_ull, assumed;
do{
assumed = old;
old = atomicCAS(address_as_ull, assumed, __double_as_longlong(val + __longlong_as_double(assumed)));
}while(assumed != old);
return __longlong_as_double(old);
}
__global__ void kernel(curandState *globalState){
// global id
int gidx = threadIdx.x + blockIdx.x * blockDim.x;
// local id
int lidx = threadIdx.x;
// creat shared memory to store seeds
__shared__ curandState localState[tot_threads];
__shared__ double srandnum[threads_per_block];
// copy global seed to local
localState[lidx] = globalState[gidx];
//synchronize the local threads writing to the local memory cache
__syncthreads();
// generate random number from normal distribution in shared memory
srandnum[lidx] = curand_normal(&localState[lidx]);
__syncthreads();
if(lidx == 0){srandnum[lidx] += srandnum[lidx + 1];} // sum of each block
if(lidx == 0){gsum[blockIdx.x] = srandnum[lidx];} // copy the sums back to global memory
__threadfence();
if( gidx < num_of_blocks){
atomicAdd_d(&dev_sum, gsum[gidx]);
}
if( gidx < num_of_blocks){
atomicMax_d(&dev_max, gsum[gidx]);
}
if( gidx == 0){
printf("Sum is: %lf\n", dev_sum);
}
if( gidx == 1){
printf("Max is: %lf\n", dev_max);
}
}
int main(){
// set seed on device
curandState *globalState;
cudaMalloc((void**)&globalState, tot_threads*sizeof(curandState));
initcuRand<<<num_of_blocks, threads_per_block>>>(globalState, 1);
// launch kernel
kernel<<<num_of_blocks, threads_per_block>>>(globalState);
double randnum[num_of_blocks];
cudaMemcpyFromSymbol(randnum, gsum, num_of_blocks*sizeof(double), 0, cudaMemcpyDeviceToHost);
std::cout << "Sum of each block:\n";
for (int i = 0; i < num_of_blocks; ++i){
std::cout << randnum[i] << std::endl;
}
cudaFree(globalState);
return 0;
}
The result I get is
Sum is: -0.898329
Max is: 0.000000
Sum of each block:
-0.0152994
-0.88303
From the result, I know that the atomicAdd function works but the atomicMax function doesn't work, I have no idea of this. Thanks beforehand.
You don't ever initialize dev_max or dev_sum. You can't sensibly do these types of atomic operations on them if they don't start with a known value.
Try something like this instead:
__device__ double dev_sum = 0.0;
__device__ double dev_max = -1e99;
and I think you'll be happier with the results.

Counting occurrences of specific events in CUDA kernels

Problem
I am trying to find the best way to count how many times my program ends up in some specific branches of my CUDA kernels. The idea is that some events should nearly never happen, but since the data processed by the GPU is given by a numerical optimization solver, there may be some situations where ill-defined cases become more common. Thus, I want to be able to track/monitor these phenomenons over multiple simulations to make some global statistics later.
Possible idea
The most straightforward way to do this may be to use a structure dedicated to monitoring such occurrences. Then, when entering a monitored branch, we increment the associated counter using atomicAdd. At the end of the simulation, we copy the counters back to the host and store them for some future statistics processing.
In my case, the cost of using atomicAdd should not be that important since I should not be entering those branches that much, but still, I may want to monitor some of the common branches later on, so what would be a better approach then? Since this is just for monitoring, I do not want the overhead to be too important.
I guess I could also have one monitoring structure per block and do a sum at the end, since it should not use much global memory anyway (1 unsigned int per monitored branch).
Code example
#include <iostream>
#include <time.h>
#include <cuda.h>
#include <stdio.h>
#define CUDA_CHECK_ERROR() __cuda_check_errors(__FILE__, __LINE__)
#define CUDA_SAFE_CALL(err) __cuda_safe_call(err, __FILE__, __LINE__)
inline void __cuda_check_errors(const char *filename, const int line_number)
{
cudaError err = cudaDeviceSynchronize();
if(err != cudaSuccess)
{
printf("CUDA error %i at %s:%i: %s\n",
err, filename, line_number, cudaGetErrorString(err));
exit(-1);
}
}
inline void __cuda_safe_call(cudaError err, const char *filename, const int line_number)
{
if (err != cudaSuccess)
{
printf("CUDA error %i at %s:%i: %s\n",
err, filename, line_number, cudaGetErrorString(err));
exit(-1);
}
}
struct Stats
{
unsigned int even;
};
__global__ void test_kernel(int* A, int* B, Stats* stats)
{
int tid = threadIdx.x + blockIdx.x * blockDim.x;
int res = A[tid] + (int)tid;
if (res%2 == 0)
atomicAdd(&(stats->even), 1);
B[tid] = res;
}
int get_random_int(int min, int max)
{
return min + (rand() % (int)(max - min + 1));
}
void print_array(int* ar, unsigned int n)
{
for (unsigned int i = 0; i < n; ++i)
std::cout << ar[i] << " ";
std::cout << std::endl;
}
void print_stats(Stats* s)
{
std::cout << "even: " << s->even << std::endl;
}
int main()
{
// vector size
const unsigned int N = 10;
// device vectors
int *d_A, *d_B;
Stats *d_stats;
// host vectors
int *h_A, *h_B;
Stats *h_stats;
// allocate device memory
CUDA_SAFE_CALL(cudaMalloc(&d_A, N * sizeof(int)));
CUDA_SAFE_CALL(cudaMalloc(&d_B, N * sizeof(int)));
CUDA_SAFE_CALL(cudaMalloc(&d_stats, sizeof(Stats)));
// allocate host memory
h_A = new int[N];
h_B = new int[N];
h_stats = new Stats;
// initialize host data
srand(time(NULL));
for (unsigned int i = 0; i < N; ++i)
{
h_A[i] = get_random_int(0,10);
h_B[i] = 0;
}
memset(h_stats, 0, sizeof(Stats));
// copy data to the device
CUDA_SAFE_CALL(cudaMemcpy(d_A, h_A, N * sizeof(int), cudaMemcpyHostToDevice));
CUDA_SAFE_CALL(cudaMemcpy(d_stats, h_stats, sizeof(Stats), cudaMemcpyHostToDevice));
// launch kernel
dim3 grid_size, block_size;
grid_size.x = N;
test_kernel<<<grid_size, block_size>>>(d_A, d_B, d_stats);
// copy result back to host
CUDA_SAFE_CALL(cudaMemcpy(h_B, d_B, N * sizeof(int), cudaMemcpyDeviceToHost));
CUDA_SAFE_CALL(cudaMemcpy(h_stats, d_stats, sizeof(Stats), cudaMemcpyDeviceToHost));
print_array(h_B, N);
print_stats(h_stats);
// free device memory
CUDA_SAFE_CALL(cudaFree(d_A));
CUDA_SAFE_CALL(cudaFree(d_B));
CUDA_SAFE_CALL(cudaFree(d_stats));
// free host memory
delete [] h_A;
delete [] h_B;
delete h_stats;
}
Hardware/software information
The solution I am looking for should work for CC >= 2.0 devices and CUDA >= 5.0.
The atomicAdd is is one possibility and i would probably go that route. If you do not use the result of the atomicAdd function call the compiler will emit a reduction operation such as RED.E.ADD. Reduction is very fast as long as there are not many conflicts happening (i actually use it sometimes even if i do not need the operation to be atomic because it can be quicker than loading value from global memory, doing an arithmetic operation and saving back to global memory).
The second option you have is to use a profiler counter and use the profiler to analyze the result. Please see Profiler Counter Function for more details.

Adding values on GPU

i have a class called Product.
Each product has a value and i want to add these values on GPU. I filled my array on host side
int * h_A, * d_A;
h_A = (int*) malloc(enterNum * sizeof(int));
cudaMalloc((void **) &d_A, enterNum * sizeof(int));
Product p("Product", price);
h_A[i] = p.getValue();
while (i < enterNum) {
i++;
cout << "Enter product name:";
cin >> desc;
cout << "Enter product price:";
cin >> price;
Product p("Product", price);
h_A[i] = p.getValue();
}
cudaMemcpy(d_A, h_A, enterNum, cudaMemcpyHostToDevice);
priceSum<<<enterNum, 1024>>>(d_A,enterNum,result);
int result2 = 0;
cudaMemcpy(result, result2, enterNum, cudaMemcpyDeviceToHost);
here cudaMemcpy function gives error because i dont use pointer. What can i do here? I dont need to use pointer here isn't it?
this is my summation function:
__global__ void priceSum(int *dA, int count, int result) {
int tid = blockIdx.x;
if (tid < count){
result+= dA[tid];
}
}
full code:
using namespace std;
#include "cuda_runtime.h"
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stdlib.h>
class Product {
private:
char * description;
int productCode;
int value;
static int lastCode;
public:
Product(char* descriptionP, int valueP) {
productCode = ++lastCode;
value = valueP;
description = new char[strlen(descriptionP) + 1];
strcpy(description, descriptionP);
}
Product(Product& other) {
productCode = ++lastCode;
description = new char[strlen(other.description) + 1];
strcpy(description, other.description);
}
~Product() {
delete[] description;
}
char* getDescription() const {
return description;
}
void setDescription(char* description) {
this->description = description;
}
int getValue() const {
return value;
}
void setValue(int value) {
this->value = value;
}
};
int Product::lastCode = 1000;
__global__ void priceSum(int *dA, int count, int * result) {
int tid = blockIdx.x;
if (tid < count)
result+= dA[tid];
}
int main(void) {
int enterNum, price, * result = 0;
string desc;
const char * desc2;
cout << "How many products do you want to enter?";
cin >> enterNum;
int * h_A, * d_A;
h_A = (int*) malloc(enterNum * sizeof(int));
cudaMalloc((void **) &d_A, enterNum * sizeof(int));
int i = 0;
while (i < enterNum) {
cout << "Enter product name:";
cin >> desc;
cout << "Enter product price:";
cin >> price;
Product p("Product", price);
h_A[i] = p.getValue();
i++;
}
cudaMemcpy(d_A, h_A, enterNum * sizeof(int), cudaMemcpyHostToDevice);
priceSum<<<enterNum, 1>>>(d_A,enterNum,result);
int result2 = 0;
cudaMemcpy(&result2, result, enterNum, cudaMemcpyDeviceToHost);
cout << result2;
return 0;
}
You should show the definition of result in your host code, but I assume it is:
int result;
based on how you are passing it to your priceSum kernel.
You have more than 1 problem here.
In your priceSum kernel, you are summing the values in dA[] and storing the answer in result. But you have passed the variable result to the kernel by value instead of by reference so the value you are modifying is local to the function, and will not show up anywhere else. When a function in C needs to modify a variable that is passed to it via the parameter list, and the modified variable is to show up in the function calling context, it's necessary to pass that parameter by reference (i.e. using a pointer) rather than by value. Note this is based on the C programming language and is not specific to CUDA. So you should rewrite your kernel definition as:
__global__ void priceSum(int *dA, int count, int *result) {
Regarding your cudaMemcpy call, there are several issues that need to be cleaned up. First, we need the storage for result to be properly created using cudaMalloc (before the kernel is called, because the kernel will store something there.) Next, we need to fix the parameter list of the cudaMemcpy call itself. So your host code should be rewritten as:
cudaMemcpy(d_A, h_A, enterNum, cudaMemcpyHostToDevice);
int *result;
cudaMalloc((void **)&result, sizeof(int));
priceSum<<<enterNum, 1024>>>(d_A,enterNum,result);
int result2 = 0;
cudaMemcpy(&result2, result, sizeof(int), cudaMemcpyDeviceToHost);
There appear to be other problems with your code, around the grouping of data for threads and blocks. But you haven't shown enough of your program for me to make sense of it. So let me point out that your code shows only a single value for result (and result2), yet the way your kernel is written, each thread will add its value of dA[tid] to result. You can't have a bunch of threads all updating a single value in global memory with no control mechanism, and expect to get a sensible result. Problems like this are usually best handled with a classical parallel reduction algorithm, but for the sake of simplicity, to try and get something working, you can use atomics:
atomicAdd(result, dA[tid]);
Sorry, but your kernel just makes no sense at all. You are using blockIdx.x as your tid variable, but let's note that blockIdx.x is a number that is the same for every thread in a particular block. So then going on to have every thread add dA[tid] to result in this fashion just doesn't make sense. I believe it will make more sense if you change your kernel invocation to:
priceSum<<<enterNum, 1>>>(d_A,enterNum,result);