How to use CUB and Thrust in one CUDA code - cuda

I'm trying to introduce some CUB into my "old" Thrust code, and so have started with a small example to compare thrust::reduce_by_key with cub::DeviceReduce::ReduceByKey, both applied to thrust::device_vectors.
The thrust part of the code is fine, but the CUB part, which naively uses raw pointers obtained via thrust::raw_pointer_cast, crashes after the CUB calls. I put in a cudaDeviceSynchronize() to try to solve this problem, but it didn't help. The CUB part of the code was cribbed from the CUB web pages.
On OSX the runtime error is:
libc++abi.dylib: terminate called throwing an exception
Abort trap: 6
On Linux the runtime error is:
terminate called after throwing an instance of 'thrust::system::system_error'
what(): an illegal memory access was encountered
The first few lines of cuda-memcheck are:
========= CUDA-MEMCHECK
========= Invalid __global__ write of size 4
========= at 0x00127010 in /home/sdettrick/codes/MCthrust/tests/../cub-1.3.2/cub/device/dispatch/../../block_range/block_range_reduce_by_key.cuh:1017:void cub::ReduceByKeyRegionKernel<cub::DeviceReduceByKeyDispatch<unsigned int*, unsigned int*, float*, float*, int*, cub::Equality, CustomSum, int>::PtxReduceByKeyPolicy, unsigned int*, unsigned int*, float*, float*, int*, cub::ReduceByKeyScanTileState<float, int, bool=1>, cub::Equality, CustomSum, int>(unsigned int*, float*, float*, int*, cub::Equality, CustomSum, int, cub::DeviceReduceByKeyDispatch<unsigned int*, unsigned int*, float*, float*, int*, cub::Equality, CustomSum, int>::PtxReduceByKeyPolicy, unsigned int*, int, cub::GridQueue<int>)
========= by thread (0,0,0) in block (0,0,0)
========= Address 0x7fff7dbb3e88 is out of bounds
========= Saved host backtrace up to driver entry point at kernel launch time
Unfortunately I'm not too sure what to do about that.
Any help would be greatly appreciated. I tried this on the NVIDIA developer zone but didn't get any responses. The complete example code is below. It should compile with CUDA 6.5 and cub 1.3.2:
#include <iostream>
#include <thrust/sort.h>
#include <thrust/gather.h>
#include <thrust/device_vector.h>
#include <thrust/iterator/zip_iterator.h>
#include <thrust/iterator/permutation_iterator.h>
#include <thrust/iterator/discard_iterator.h>
#include <cub/cub.cuh> // or equivalently <cub/device/device_radix_sort.cuh>
//========================================
// for CUB:
struct CustomSum
{
template <typename T>
CUB_RUNTIME_FUNCTION __host__ __device__ __forceinline__
//__host__ __device__ __forceinline__
T operator()(const T &a, const T &b) const {
return b+a;
}
};
//========================================
int main()
{
const int Nkey=20;
int Nseg=9;
int ikey[Nkey] = {0, 0, 0, 6, 8, 0, 2, 4, 6, 8, 1, 3, 5, 7, 8, 1, 3, 5, 7, 8};
thrust::device_vector<unsigned int> key(ikey,ikey+Nkey);
thrust::device_vector<unsigned int> keysout(Nkey);
// Let's reduce x, by key:
float xval[Nkey];
for (int i=0; i<Nkey; i++) xval[i]=ikey[i]+0.1f;
thrust::device_vector<float> x(xval,xval+Nkey);
// First, sort x by key:
thrust::sort_by_key(key.begin(),key.end(),x.begin());
//---------------------------------------------------------------------
std::cout<<"=================================================================="<<std::endl
<<" THRUST reduce_by_key:"<<std::endl
<<"=================================================================="<<std::endl;
thrust::device_vector<float> output(Nseg,0.0f);
thrust::reduce_by_key(key.begin(),
key.end(),
x.begin(),
keysout.begin(),
output.begin());
for (int i=0;i<Nkey;i++) std::cout << x[i] <<" "; std::cout<<std::endl;
for (int i=0;i<Nkey;i++) std::cout << key[i] <<" "; std::cout<<std::endl;
for (int i=0;i<Nseg;i++) std::cout << output[i] <<" "; std::cout<<std::endl;
float ototal=thrust::reduce(output.begin(),output.end());
float xtotal=thrust::reduce(x.begin(),x.end());
std::cout << "total="<< ototal <<", should be "<<xtotal<<std::endl;
//---------------------------------------------------------------------
std::cout<<"=================================================================="<<std::endl
<<" CUB ReduceByKey:"<<std::endl
<<"=================================================================="<<std::endl;
unsigned int *d_keys_in =thrust::raw_pointer_cast(&key[0]);
float *d_values_in =thrust::raw_pointer_cast(&x[0]);
unsigned int *d_keys_out =thrust::raw_pointer_cast(&keysout[0]);
float *d_values_out=thrust::raw_pointer_cast(&output[0]);
int *d_num_segments=&Nseg;
CustomSum reduction_op;
std::cout << "CUB input" << std::endl;
for (int i=0; i<Nkey; ++i) std::cout << key[i] << " "; std::cout<<std::endl;
for (int i=0; i<Nkey; ++i) std::cout << x[i] << " "; std::cout<< std::endl;
for (int i=0; i<Nkey; ++i) std::cout << keysout[i] << " "; std::cout<< std::endl;
for (int i=0; i<Nseg; ++i) std::cout << output[i] << " "; std::cout<< std::endl;
// Determine temporary device storage requirements
void *d_temp_storage = NULL;
size_t temp_storage_bytes = 0;
cub::DeviceReduce::ReduceByKey(d_temp_storage, temp_storage_bytes, d_keys_in, d_keys_out, d_values_in, d_values_out, d_num_segments, reduction_op, Nkey);
// Allocate temporary storage
cudaMalloc(&d_temp_storage, temp_storage_bytes);
std::cout << "temp_storage_bytes = " << temp_storage_bytes << std::endl;
// Run reduce-by-key
cub::DeviceReduce::ReduceByKey(d_temp_storage, temp_storage_bytes, d_keys_in, d_keys_out, d_values_in, d_values_out, d_num_segments, reduction_op, Nkey);
cudaDeviceSynchronize();
std::cout << "CUB output" << std::endl;
std::cout<<Nkey<<" "<<Nseg<<std::endl;
std::cout<<key.size() << " "<<x.size() << " "<<keysout.size() << " "<<output.size() << std::endl;
// At this point onward it dies:
//libc++abi.dylib: terminate called throwing an exception
//Abort trap: 6
// If the next line is uncommented, it crashes the Mac!
for (int i=0; i<Nkey; ++i) std::cout << key[i] << " "; std::cout<<std::endl;
// for (int i=0; i<Nkey; ++i) std::cout << x[i] << " "; std::cout<< std::endl;
// for (int i=0; i<Nkey; ++i) std::cout << keysout[i] << " "; std::cout<< std::endl;
// for (int i=0; i<Nseg; ++i) std::cout << output[i] << " "; std::cout<< std::endl;
cudaFree(d_temp_storage);
ototal=thrust::reduce(output.begin(),output.end());
xtotal=thrust::reduce(x.begin(),x.end());
std::cout << "total="<< ototal <<", should be "<<xtotal<<std::endl;
return 1;
}

This is not appropriate:
int *d_num_segments=&Nseg;
You cannot take the address of a host variable and use it as a device pointer.
Instead do this:
int *d_num_segments;
cudaMalloc(&d_num_segments, sizeof(int));
This allocates space on the device for the size of data (a single integer that cub will write to), and assigns the address of that allocation to your d_num_segments variable. This then becomes a valid device pointer.
In (*ordinary, non-UM) CUDA, it is illegal dereference a host address in device code, or a device address in host code.

Related

How to cope with "cudaErrorMissingConfiguration" from "cudaMallocPitch" function of CUDA?

I'm making a Mandelbrot set program with CUDA. However I can't step more unless cudaErrorMissingConfiguration from cudaMallocPitch() function of CUDA is to be solved. Could you tell me something about it?
My GPU is GeForce RTX 2060 SUPER.
I'll show you my command lines below.
> nvcc MandelbrotCUDA.cu -o MandelbrotCUDA -O3
I tried cudaDeviceSetLimit( cudaLimitMallocHeapSize, 7*1024*1024*1024 ) to
resize heap size.
cudaDeviceSetLimit was success.
However I cannot step one more. I cannot print "CUDA malloc done!"
#include <iostream>
#include <thrust/complex.h>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;
#define D 0.0000025 // Tick
#define LIMIT_N 255
#define INF_NUM 2
#define PLOT_METHOD 2 // dat file : 0, ppm file : 1, ppm file with C : 2
__global__
void calculation(const int indexTotalX, const int indexTotalY, int ***n, thrust::complex<double> ***c){ // n, c are the pointers of dN, dC.
for(int i = 0; i < indexTotalY ; i++){
for(int j = 0; j < indexTotalX; j++){
thrust::complex<double> z(0.0f, 0.0f);
n[i][j] = 0;
for(int ctr=1; ctr <= LIMIT_N ; ctr++){
z = z*z + (*(c[i][j]));
n[i][j] = n[i][j] + (abs(z) < INF_NUM);
}
}
}
}
int main(){
// Data Path
string filePath = "Y:\\Documents\\Programming\\mandelbrot\\";
string fileName = "mandelbrot4.ppm";
string filename = filePath+fileName;
//complex<double> c[N][M];
double xRange[2] = {-0.76, -0.74};
double yRange[2] = {0.05, 0.1};
const int indexTotalX = (xRange[1]-xRange[0])/D;
const int indexTotalY = (yRange[1]-yRange[0])/D;
thrust::complex<double> **c;
//c = new complex<double> [N];
cout << "debug_n" << endl;
int **n;
n = new int* [indexTotalY];
c = new thrust::complex<double> * [indexTotalY];
for(int i=0;i<indexTotalY;i++){
n[i] = new int [indexTotalX];
c[i] = new thrust::complex<double> [indexTotalX];
}
cout << "debug_n_end" << endl;
for(int i = 0; i < indexTotalY; i++){
for(int j = 0; j < indexTotalX; j++){
thrust::complex<double> tmp( xRange[0]+j*D, yRange[0]+i*D );
c[i][j] = tmp;
//n[i*sqrt(N)+j] = 0;
}
}
// CUDA malloc
cout << "CUDA malloc initializing..." << endl;
int **dN;
thrust::complex<double> **dC;
cudaError_t error;
error = cudaDeviceSetLimit(cudaLimitMallocHeapSize, 7*1024*1024*1024);
if(error != cudaSuccess){
cout << "cudaDeviceSetLimit's ERROR CODE = " << error << endl;
return 0;
}
size_t tmpPitch;
error = cudaMallocPitch((void **)dN, &tmpPitch,(size_t)(indexTotalY*sizeof(int)), (size_t)(indexTotalX*sizeof(int)));
if(error != cudaSuccess){
cout << "CUDA ERROR CODE = " << error << endl;
cout << "indexTotalX = " << indexTotalX << endl;
cout << "indexTotalY = " << indexTotalY << endl;
return 0;
}
cout << "CUDA malloc done!" << endl;
This is console messages below.
debug_n
debug_n_end
CUDA malloc initializing...
CUDA ERROR CODE = 1
indexTotalX = 8000
indexTotalY = 20000
There are several problems here:
int **dN;
...
error = cudaMallocPitch((void **)dN, &tmpPitch,(size_t)(indexTotalY*sizeof(int)), (size_t)(indexTotalX*sizeof(int)));
The correct type of pointer to use in CUDA allocations is a single pointer:
int *dN;
not a double pointer:
int **dN;
(so your kernel where you are trying pass triple-pointers:
void calculation(const int indexTotalX, const int indexTotalY, int ***n, thrust::complex<double> ***c){ // n, c are the pointers of dN, dC.
is almost certainly not going to work, and should not be designed that way, but that is not the question you are asking.)
The pointer is passed to the allocating function by its address:
error = cudaMallocPitch((void **)&dN,
For cudaMallocPitch, only the horizontal requested dimension is scaled by the size of the data element. The allocation height is not scaled this way. Also, I will assume X corresponds to your allocation width, and Y corresponds to your allocation height, so you also have those parameters reversed:
error = cudaMallocPitch((void **)&dN, &tmpPitch,(size_t)(indexTotalX*sizeof(int)), (size_t)(indexTotalY));
The cudaLimitMallocHeapSize should not be necessary to set to make any of this work. It applies only to in-kernel allocations. Reserving 7GB on an 8GB card may also cause problems. Until you are sure you need that (it's not needed for what you have shown) I would simply remove that.
$ cat t1488.cu
#include <iostream>
#include <thrust/complex.h>
#include <fstream>
#include <string>
#include <stdlib.h>
using namespace std;
#define D 0.0000025 // Tick
#define LIMIT_N 255
#define INF_NUM 2
#define PLOT_METHOD 2 // dat file : 0, ppm file : 1, ppm file with C : 2
__global__
void calculation(const int indexTotalX, const int indexTotalY, int ***n, thrust::complex<double> ***c){ // n, c are the pointers of dN, dC.
for(int i = 0; i < indexTotalY ; i++){
for(int j = 0; j < indexTotalX; j++){
thrust::complex<double> z(0.0f, 0.0f);
n[i][j] = 0;
for(int ctr=1; ctr <= LIMIT_N ; ctr++){
z = z*z + (*(c[i][j]));
n[i][j] = n[i][j] + (abs(z) < INF_NUM);
}
}
}
}
int main(){
// Data Path
string filePath = "Y:\\Documents\\Programming\\mandelbrot\\";
string fileName = "mandelbrot4.ppm";
string filename = filePath+fileName;
//complex<double> c[N][M];
double xRange[2] = {-0.76, -0.74};
double yRange[2] = {0.05, 0.1};
const int indexTotalX = (xRange[1]-xRange[0])/D;
const int indexTotalY = (yRange[1]-yRange[0])/D;
thrust::complex<double> **c;
//c = new complex<double> [N];
cout << "debug_n" << endl;
int **n;
n = new int* [indexTotalY];
c = new thrust::complex<double> * [indexTotalY];
for(int i=0;i<indexTotalY;i++){
n[i] = new int [indexTotalX];
c[i] = new thrust::complex<double> [indexTotalX];
}
cout << "debug_n_end" << endl;
for(int i = 0; i < indexTotalY; i++){
for(int j = 0; j < indexTotalX; j++){
thrust::complex<double> tmp( xRange[0]+j*D, yRange[0]+i*D );
c[i][j] = tmp;
//n[i*sqrt(N)+j] = 0;
}
}
// CUDA malloc
cout << "CUDA malloc initializing..." << endl;
int *dN;
thrust::complex<double> **dC;
cudaError_t error;
size_t tmpPitch;
error = cudaMallocPitch((void **)&dN, &tmpPitch,(size_t)(indexTotalX*sizeof(int)), (size_t)(indexTotalY));
if(error != cudaSuccess){
cout << "CUDA ERROR CODE = " << error << endl;
cout << "indexTotalX = " << indexTotalX << endl;
cout << "indexTotalY = " << indexTotalY << endl;
return 0;
}
cout << "CUDA malloc done!" << endl;
}
$ nvcc -o t1488 t1488.cu
t1488.cu(68): warning: variable "dC" was declared but never referenced
$ cuda-memcheck ./t1488
========= CUDA-MEMCHECK
debug_n
debug_n_end
CUDA malloc initializing...
CUDA malloc done!
========= ERROR SUMMARY: 0 errors
$

False dependency issue for the Fermi architecture

I am trying to achieve "3-way overlapping" using 3 streams as in the examples in CUDA streams and concurrency webinar. But I couldn't achieve it.
I have Geforce GT 550M (Fermi Architecture with one copy engine) and I am using Windows 7 (64 bit).
Here is the code that I have written.
#include <iostream>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
// includes, project
#include "helper_cuda.h"
#include "helper_functions.h" // helper utility functions
#include <stdio.h>
using namespace std;
#define DATA_SIZE 6000000
#define NUM_THREADS 32
#define NUM_BLOCKS 16
#define NUM_STREAMS 3
__global__ void kernel(const int *in, int *out, int dataSize)
{
int start = blockIdx.x * blockDim.x + threadIdx.x;
int end = dataSize;
for (int i = start; i < end; i += blockDim.x * gridDim.x)
{
out[i] = in[i] * in[i];
}
}
int main()
{
const int dataSize = DATA_SIZE;
int *h_in = new int[dataSize];
int *h_out = new int[dataSize];
int *h_groundTruth = new int[dataSize];
// Input population
for(int i = 0; i < dataSize; i++)
h_in[i] = 5;
for(int i = 0; i < dataSize; i++)
h_out[i] = 0;
// CPU calculation for ground truth
for(int i = 0; i < dataSize; i++)
h_groundTruth[i] = h_in[i] * h_in[i];
// Choose which GPU to run on, change this on a multi-GPU system.
checkCudaErrors( cudaSetDevice(0) );
int *d_in = 0;
int *d_out = 0;
int streamSize = dataSize / NUM_STREAMS;
size_t memSize = dataSize * sizeof(int);
size_t streamMemSize = memSize / NUM_STREAMS;
checkCudaErrors( cudaMalloc( (void **)&d_in, memSize) );
checkCudaErrors( cudaMalloc( (void **)&d_out, memSize) );
// registers host memory as page-locked (required for asynch cudaMemcpyAsync)
checkCudaErrors(cudaHostRegister(h_in, memSize, cudaHostRegisterPortable));
checkCudaErrors(cudaHostRegister(h_out, memSize, cudaHostRegisterPortable));
// set kernel launch config
dim3 nThreads = dim3(NUM_THREADS,1,1);
dim3 nBlocks = dim3(NUM_BLOCKS,1,1);
cout << "GPU Kernel Configuration : " << endl;
cout << "Number of Streams :\t" << NUM_STREAMS << " with size: \t" << streamSize << endl;
cout << "Number of Threads :\t" << nThreads.x << "\t" << nThreads.y << "\t" << nThreads.z << endl;
cout << "Number of Blocks :\t" << nBlocks.x << "\t" << nBlocks.y << "\t" << nBlocks.z << endl;
// create cuda stream
cudaStream_t streams[NUM_STREAMS];
for(int i = 0; i < NUM_STREAMS; i++)
checkCudaErrors(cudaStreamCreate(&streams[i]));
// create cuda event handles
cudaEvent_t start, stop;
checkCudaErrors(cudaEventCreate(&start));
checkCudaErrors(cudaEventCreate(&stop));
cudaEventRecord(start, 0);
// overlapped execution using version 2
for(int i = 0; i < NUM_STREAMS; i++)
{
int offset = i * streamSize;
cudaMemcpyAsync(&d_in[offset], &h_in[offset], streamMemSize, cudaMemcpyHostToDevice, streams[i]);
}
//cudaMemcpy(d_in, h_in, memSize, cudaMemcpyHostToDevice);
for(int i = 0; i < NUM_STREAMS; i++)
{
int offset = i * streamSize;
dim3 subKernelBlock = dim3((int)ceil((float)nBlocks.x / 2));
//kernel<<<nBlocks, nThreads, 0, streams[i]>>>(&d_in[offset], &d_out[offset], streamSize);
kernel<<<subKernelBlock, nThreads, 0, streams[i]>>>(&d_in[offset], &d_out[offset], streamSize/2);
kernel<<<subKernelBlock, nThreads, 0, streams[i]>>>(&d_in[offset + streamSize/2], &d_out[offset + streamSize/2], streamSize/2);
}
for(int i = 0; i < NUM_STREAMS; i++)
{
int offset = i * streamSize;
cudaMemcpyAsync(&h_out[offset], &d_out[offset], streamMemSize, cudaMemcpyDeviceToHost, streams[i]);
}
for(int i = 0; i < NUM_STREAMS; i++)
checkCudaErrors(cudaStreamSynchronize(streams[i]));
cudaEventRecord(stop, 0);
checkCudaErrors(cudaStreamSynchronize(0));
checkCudaErrors(cudaDeviceSynchronize());
float gpu_time = 0;
checkCudaErrors(cudaEventElapsedTime(&gpu_time, start, stop));
// release resources
checkCudaErrors(cudaEventDestroy(start));
checkCudaErrors(cudaEventDestroy(stop));
checkCudaErrors(cudaHostUnregister(h_in));
checkCudaErrors(cudaHostUnregister(h_out));
checkCudaErrors(cudaFree(d_in));
checkCudaErrors(cudaFree(d_out));
for(int i = 0; i < NUM_STREAMS; i++)
checkCudaErrors(cudaStreamDestroy(streams[i]));
cudaDeviceReset();
cout << "Execution Time of GPU: " << gpu_time << "ms" << endl;
// GPU output check
int sum = 0;
for(int i = 0; i < dataSize; i++)
sum += h_groundTruth[i] - h_out[i];
cout << "Error between CPU and GPU: " << sum << endl;
delete[] h_in;
delete[] h_out;
delete[] h_groundTruth;
return 0;
}
Using Nsight for profiling, I have this result:
It may seem correct, but why does the D2H transfer in stream #1 only start when the last kernel launch of stream #2 and not before?
I tried also to use 8 streams (just by changing NUM_STREAM to 8) to achieve such a "3-way overlap" and here is the result:
The interesting thing is that when I use 8 streams, the overlappings between computation and memory transfers seem to be much better.
What is the reason for this problem? Is it due to WDDM driver or is there something wrong with my program?
From the comments above, it seems that the OP's problem is a false dependency issue, suffered by the Fermi architecture and solved by the Hyper-Q feature of the Kepler architecture.
To summarize, the OP is highlighting the fact that the first D2H transfer (stream #1) does not start immediately after the last H2D (stream #3) finishes, while in principle it could. The time gap is highlighted by the red circle in the following figure (henceforth, but for the differently specified, all the tests refer to a GeForce GT540M belonging to the Fermi family):
The OP's approach is a breadth-first approach, which operates according to the following scheme:
for(int i = 0; i < NUM_STREAMS; i++)
cudaMemcpyAsync(..., cudaMemcpyHostToDevice, streams[i]);
for(int i = 0; i < NUM_STREAMS; i++)
{
kernel_launch_1<<<..., 0, streams[i]>>>(...);
kernel_launch_2<<<..., 0, streams[i]>>>(...);
}
for(int i = 0; i < NUM_STREAMS; i++)
cudaMemcpyAsync(..., cudaMemcpyDeviceToHost, streams[i]);
Using a depth-first approach, operating according to the following scheme
for(int i = 0; i < NUM_STREAMS; i++)
{
cudaMemcpyAsync(...., cudaMemcpyHostToDevice, streams[i]);
kernel_launch_1<<<...., 0, streams[i]>>>(....);
kernel_launch_2<<<...., 0, streams[i]>>>(....);
cudaMemcpyAsync(...., cudaMemcpyDeviceToHost, streams[i]);
}
does not seem to improve the situation, according to the following timeline (the depth-first code is reported at the bottom of the answer), but it seems to show a worse overlapping:
Under the breadth-first approach, and commenting the second kernel launch, the first D2H copy starts immediately as it can, as reported by the following timeline:
Finally, running the code on a Kepler K20c, the problem does not show up, as illustrated by the following figure:
Here is the code for the depth-first approach:
#include <iostream>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
// includes, project
#include "helper_cuda.h"
#include "helper_functions.h" // helper utility functions
#include <stdio.h>
using namespace std;
#define DATA_SIZE 6000000
#define NUM_THREADS 32
#define NUM_BLOCKS 16
#define NUM_STREAMS 3
__global__ void kernel(const int *in, int *out, int dataSize)
{
int start = blockIdx.x * blockDim.x + threadIdx.x;
int end = dataSize;
for (int i = start; i < end; i += blockDim.x * gridDim.x)
{
out[i] = in[i] * in[i];
}
}
int main()
{
const int dataSize = DATA_SIZE;
int *h_in = new int[dataSize];
int *h_out = new int[dataSize];
int *h_groundTruth = new int[dataSize];
// Input population
for(int i = 0; i < dataSize; i++)
h_in[i] = 5;
for(int i = 0; i < dataSize; i++)
h_out[i] = 0;
// CPU calculation for ground truth
for(int i = 0; i < dataSize; i++)
h_groundTruth[i] = h_in[i] * h_in[i];
// Choose which GPU to run on, change this on a multi-GPU system.
checkCudaErrors( cudaSetDevice(0) );
int *d_in = 0;
int *d_out = 0;
int streamSize = dataSize / NUM_STREAMS;
size_t memSize = dataSize * sizeof(int);
size_t streamMemSize = memSize / NUM_STREAMS;
checkCudaErrors( cudaMalloc( (void **)&d_in, memSize) );
checkCudaErrors( cudaMalloc( (void **)&d_out, memSize) );
// registers host memory as page-locked (required for asynch cudaMemcpyAsync)
checkCudaErrors(cudaHostRegister(h_in, memSize, cudaHostRegisterPortable));
checkCudaErrors(cudaHostRegister(h_out, memSize, cudaHostRegisterPortable));
// set kernel launch config
dim3 nThreads = dim3(NUM_THREADS,1,1);
dim3 nBlocks = dim3(NUM_BLOCKS,1,1);
cout << "GPU Kernel Configuration : " << endl;
cout << "Number of Streams :\t" << NUM_STREAMS << " with size: \t" << streamSize << endl;
cout << "Number of Threads :\t" << nThreads.x << "\t" << nThreads.y << "\t" << nThreads.z << endl;
cout << "Number of Blocks :\t" << nBlocks.x << "\t" << nBlocks.y << "\t" << nBlocks.z << endl;
// create cuda stream
cudaStream_t streams[NUM_STREAMS];
for(int i = 0; i < NUM_STREAMS; i++)
checkCudaErrors(cudaStreamCreate(&streams[i]));
// create cuda event handles
cudaEvent_t start, stop;
checkCudaErrors(cudaEventCreate(&start));
checkCudaErrors(cudaEventCreate(&stop));
cudaEventRecord(start, 0);
for(int i = 0; i < NUM_STREAMS; i++)
{
int offset = i * streamSize;
cudaMemcpyAsync(&d_in[offset], &h_in[offset], streamMemSize, cudaMemcpyHostToDevice, streams[i]);
dim3 subKernelBlock = dim3((int)ceil((float)nBlocks.x / 2));
kernel<<<subKernelBlock, nThreads, 0, streams[i]>>>(&d_in[offset], &d_out[offset], streamSize/2);
kernel<<<subKernelBlock, nThreads, 0, streams[i]>>>(&d_in[offset + streamSize/2], &d_out[offset + streamSize/2], streamSize/2);
cudaMemcpyAsync(&h_out[offset], &d_out[offset], streamMemSize, cudaMemcpyDeviceToHost, streams[i]);
}
for(int i = 0; i < NUM_STREAMS; i++)
checkCudaErrors(cudaStreamSynchronize(streams[i]));
cudaEventRecord(stop, 0);
checkCudaErrors(cudaStreamSynchronize(0));
checkCudaErrors(cudaDeviceSynchronize());
float gpu_time = 0;
checkCudaErrors(cudaEventElapsedTime(&gpu_time, start, stop));
// release resources
checkCudaErrors(cudaEventDestroy(start));
checkCudaErrors(cudaEventDestroy(stop));
checkCudaErrors(cudaHostUnregister(h_in));
checkCudaErrors(cudaHostUnregister(h_out));
checkCudaErrors(cudaFree(d_in));
checkCudaErrors(cudaFree(d_out));
for(int i = 0; i < NUM_STREAMS; i++)
checkCudaErrors(cudaStreamDestroy(streams[i]));
cudaDeviceReset();
cout << "Execution Time of GPU: " << gpu_time << "ms" << endl;
// GPU output check
int sum = 0;
for(int i = 0; i < dataSize; i++)
sum += h_groundTruth[i] - h_out[i];
cout << "Error between CPU and GPU: " << sum << endl;
delete[] h_in;
delete[] h_out;
delete[] h_groundTruth;
return 0;
}

reduction example using cuda and CUB

I'm trying to get my head around CUB, and having a bit of trouble following the (rather incomplete) worked examples. CUB looks like it is a fantastic tool, I just can't make sense of the example code.
I've built a simple proto-warp reduce example:
#include <cub/cub.cuh>
#include <cuda.h>
#include <vector>
using std::vector;
#include <iostream>
using std::cout;
using std::endl;
const int N = 128;
__global__ void sum(float *indata, float *outdata) {
typedef cub::WarpReduce<float,4> WarpReduce;
__shared__ typename WarpReduce::TempStorage temp_storage;
int id = blockIdx.x*blockDim.x+threadIdx.x;
if( id < 128 ) {
outdata[id] = WarpReduce(temp_storage).Sum(indata[id]);
}
}
int main() {
vector<float> y(N), sol(N);
float *dev_y, *dev_sol;
cudaMalloc((void**)&dev_y,N*sizeof(float));
cudaMalloc((void**)&dev_sol,N*sizeof(float));
for( int i = 0; i < N; i++ ) {
y[i] = (float)i;
}
cout << "input: ";
for( int i = 0; i < N; i++ ) cout << y[i] << " ";
cout << endl;
cudaMemcpy(&y[0],dev_y,N*sizeof(float),cudaMemcpyHostToDevice);
sum<<<1,32>>>(dev_y,dev_sol);
cudaMemcpy(dev_sol,&sol[0],N*sizeof(float),cudaMemcpyDeviceToHost);
cout << "output: ";
for( int i = 0; i < N; i++ ) cout << sol[i] << " ";
cout << endl;
cudaFree(dev_y);
cudaFree(dev_sol);
return 0;
}
which returns all zeros.
I'm aware that this code would return a reduction that was banded with every 32nd element being the sum of a warp and the other elements being undefined - I just want to get a feel for how CUB works. Can someone point out what I'm doing wrong?
(also, does CUB deserve its own tag yet?)
Your cudaMemcpy arguments are back to front, the destination comes first (to be consistent with memcpy).
cudaError_t cudaMemcpy ( void* dst, const void* src, size_t count, cudaMemcpyKind kind )
See the API reference for more info.

Cuda Reduction in 2d Array

I want to calculate the average of the values over the whole image in Cuda. To test how reduction in 2D array work, I write this kernel below. The final output o should be the sum of all the image values. The input g is a 2D array with value 1 in every pixel. But the result of this program is 0 as the sum. A bit weird to me.
I imitate the reduction in 1D array in this tutorial http://developer.download.nvidia.com/compute/cuda/1.1-Beta/x86_website/projects/reduction/doc/reduction.pdf I write this 2D form. I am new to Cuda. And suggestions to potential bugs and improvement are welcomed!
Just add one comment. I know it makes sense just to calculate the average in 1D array. But I want to exploit more and test more complicated reduction behaviours. It might not be right. But just a test. Hope anyone can give me suggestions more about reduction common practices.
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
cudaEvent_t start, stop;
float elapsedTime;
__global__ void
reduce(float *g, float *o, const int dimx, const int dimy)
{
extern __shared__ float sdata[];
unsigned int tid_x = threadIdx.x;
unsigned int tid_y = threadIdx.y;
unsigned int i = blockDim.x * blockIdx.x + threadIdx.x;
unsigned int j = blockDim.y * blockIdx.y + threadIdx.y;
if (i >= dimx || j >= dimy)
return;
sdata[tid_x*blockDim.y + tid_y] = g[i*dimy + j];
__syncthreads();
for(unsigned int s_y = blockDim.y/2; s_y > 0; s_y >>= 1)
{
if (tid_y < s_y)
{
sdata[tid_x * dimy + tid_y] += sdata[tid_x * dimy + tid_y + s_y];
}
__syncthreads();
}
for(unsigned int s_x = blockDim.x/2; s_x > 0; s_x >>= 1 )
{
if(tid_x < s_x)
{
sdata[tid_x * dimy] += sdata[(tid_x + s_x) * dimy];
}
__syncthreads();
}
float sum;
if( tid_x == 0 && tid_y == 0)
{
sum = sdata[0];
atomicAdd (o, sum); // The result should be the sum of all pixel values. But the program produce 0
}
//if(tid_x==0 && tid__y == 0 )
//o[blockIdx.x] = sdata[0];
}
int
main()
{
int dimx = 320;
int dimy = 160;
int num_bytes = dimx*dimy*sizeof(float);
float *d_a, *h_a, // device and host pointers
*d_o=0, *h_o=0;
h_a = (float*)malloc(num_bytes);
h_o = (float*)malloc(sizeof(float));
srand(time(NULL));
for (int i=0; i < dimx; i++)
{
for (int j=0; j < dimy; j++)
{
h_a[i*dimy + j] = 1;
}
}
cudaMalloc( (void**)&d_a, num_bytes );
cudaMalloc( (void**)&d_o, sizeof(int) );
cudaMemcpy( d_a, h_a, num_bytes, cudaMemcpyHostToDevice);
cudaMemcpy( d_o, h_o, sizeof(int), cudaMemcpyHostToDevice);
dim3 grid, block;
block.x = 4;
block.y = 4;
grid.x = dimx / block.x;
grid.y = dimy / block.y;
cudaEventCreate(&start);
cudaEventRecord(start, 0);
int sizeofSharedMemory = dimx*dimy*sizeof(float);
reduce<<<grid, block, sizeofSharedMemory>>> (d_a, d_o, block.x, block.y);
cudaEventCreate(&stop);
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
cudaEventElapsedTime(&elapsedTime, start, stop);
std::cout << "This kernel runs: " << elapsedTime << "ms" << std::endl;
std::cout << block.x << " " << block.y << std::endl;
std::cout << grid.x << " " << grid.y << std::endl;
std::cout << dimx << " " << dimy << " " << dimx*dimy << std::endl;
cudaMemcpy( h_a, d_a, num_bytes, cudaMemcpyDeviceToHost );
cudaMemcpy( h_o, d_o, sizeof(int), cudaMemcpyDeviceToHost );
std::cout << "The sum is:" << *h_o << std::endl;
free(h_a);
free(h_o);
cudaFree(d_a);
cudaFree(d_o);
}
If you do basic cuda error checking you will discover that your reduce kernel is not even running. The reason is as follows:
int dimx = 320;
int dimy = 160;
...
int sizeofSharedMemory = dimx*dimy*sizeof(float); // = 204800
reduce<<<grid, block, sizeofSharedMemory>>> (d_a, d_o, block.x, block.y);
^
|
204800 is illegal here
You cannot request 204800 bytes of shared memory dynamically (or any other way). The maximum is slightly less than 48K bytes.
If you had done proper cuda error checking, you would discover your kernel is not running and would have gotten an instructive error message which suggests the launch configuration (the numbers between the <<< ... >>> ) is invalid. Shared memory is requested on a per-block basis, and it's probably not sensible that you need to request enough shared memory to cover your entire 2D data set, when each block only consists of a 4x4 thread array. You probably just need enough data for what will be accessed by each 4x4 thread array.
After you have properly instrumented your code with cuda error checking, and detected and corrected all the errors, then run your code with cuda-memcheck. This will do an additional level of error checking to point out any kernel access errors. You may also use cuda-memcheck if you are getting an unspecified launch failure, and it may help pinpoint the issue.
After you have done these basic trouble shooting steps, then it might make sense to ask others for help. But use the power of the tools you have been given first.
I also want to point out one other error before you come back and post this code again, asking for help.
This will not be useful:
std::cout << "The sum is:" << *h_o << std::endl;
cudaMemcpy( h_a, d_a, num_bytes, cudaMemcpyDeviceToHost );
cudaMemcpy( h_o, d_o, sizeof(int), cudaMemcpyDeviceToHost );
You are printing out the sum before you have copied the sum from the device to the host.
Reverse the order of these steps:
cudaMemcpy( h_a, d_a, num_bytes, cudaMemcpyDeviceToHost );
cudaMemcpy( h_o, d_o, sizeof(int), cudaMemcpyDeviceToHost );
std::cout << "The sum is:" << *h_o << std::endl;

CUBLAS works unpredictably

Wrote my first program using CUDA+CUBLAS. It just uses a 'cublasDgemm' function and computes a product of 2 N*N matrices.
However, all the time I was launching my program, it keeped producing the same wrong answer (e.g. when multiplying 1*1 matrix containing 5 as a single element by 1*1 matrix containing element 6, it always said the result is 36, not 30).
I checked the program several times with no success. But, when I came back to it the nexy day (i.e. after reboot), it worked just fine. I don't remember whether I recompiled it or not, but the truth is that it is the same VS project, same code, same computer with its GPU.
So, can anyone explain me why could that have happened? And do I have to expect same strange behaviour further?
Here is the code I was launching:
#include <iostream>
#include <string>
#include <iomanip>
#include <cuda_runtime.h>
#include <cublas_v2.h>
const int N = 5;
#define IDX2F(i,j) ((i) * N + j)
void fail(const cudaError_t& cudaStatus, const std::string& errorMessage) {
if (cudaStatus != cudaSuccess) {
std::cerr << errorMessage << std::endl;
exit(EXIT_FAILURE);
}
}
void fail(const cublasStatus_t& status, const std::string& errorMessage) {
if (status != CUBLAS_STATUS_SUCCESS) {
std::cerr << errorMessage << std::endl;
exit(EXIT_FAILURE);
}
}
void printMatrix(const double *C) {
for (int i=0; i<N; i++) {
for (int j=0; j<N; j++) {
std::cout << std::fixed << std::setprecision(2) << C[IDX2F(i,j)] << ' ';
}
std::cout << std::endl;
}
std::cout << std::endl;
}
int main(int argc, char **argv) {
cudaError_t cudaStatus;
cublasStatus_t status;
cublasHandle_t handle;
double *A = new double[N*N];
double *devPtrA;
double *B = new double[N*N];
double *devPtrB;
double *C = new double[N*N];
double *devPtrC;
for (int i=0; i<N; i++)
for (int j=0; j<N; j++)
A[IDX2F(i,j)] = i + j;
for (int i=0; i<N; i++)
for (int j=0; j<N; j++)
B[IDX2F(i,j)] = i + j * 0.5;
// do not have to set anything into matrix C, because beta = 0
// allocate mamory on GPU
cudaStatus = cudaMalloc((void**)&devPtrC, N*N*sizeof(*C));
fail(cudaStatus, "device memory allocation failed");
cudaStatus = cudaMalloc((void**)&devPtrA, N*N*sizeof(*A));
fail(cudaStatus, "device memory allocation failed");
cudaStatus = cudaMalloc((void**)&devPtrB, N*N*sizeof(*B));
fail(cudaStatus, "device memory allocation failed");
// create GPU handle
status = cublasCreate(&handle);
fail(status, "CUBLAS initialization failed");
// copying matrices from host to GPU
status = cublasSetMatrix(N, N, sizeof (*B), B, N, devPtrB, N);
fail(status, "failed to load data from host to GPU");
status = cublasSetMatrix(N, N, sizeof (*A), A, N, devPtrA, N);
fail(status, "failed to load data from host to GPU");
const double ONE = 1;
const double ZERO = 0;
printMatrix(A);
printMatrix(B);
status = cublasDgemm( handle,
CUBLAS_OP_N, CUBLAS_OP_N,
N, N, N,
&ONE,
devPtrA, N,
devPtrB, N,
&ZERO,
devPtrC, N);
fail(status, "error cublasDgemm");
status = cublasGetMatrix(N, N, sizeof (*C), devPtrC, N, C, N);
fail(status, "could not load result back from GPU to host");
printMatrix(C);
status = cublasDestroy(handle);
fail(status, "could not destroy CUBLAS handle");
cudaStatus = cudaFree(devPtrC);
fail(cudaStatus, "device memory freeing failed");
cudaStatus = cudaFree(devPtrB);
fail(cudaStatus, "device memory freeing failed");
cudaStatus = cudaFree(devPtrA);
fail(cudaStatus, "device memory freeing failed");
delete[] C;
delete[] B;
delete[] A;
return EXIT_SUCCESS;
}
op(B) must be CUBLAS_OP_T
.
.
status = cublasDgemm( handle,
CUBLAS_OP_N, CUBLAS_OP_T,
N, N, N,
&ONE,
devPtrA, N,
devPtrB, N,
&ZERO,
devPtrC, N);
.
.
.
.
definition is : C = α op ( A ) op ( B ) + β C
http://docs.nvidia.com/cuda/cublas/index.html#topic_8_1