OpenCL: Can one kernel call the other kernel - opencl

Hi ,
I am trying to run the available convolution code in OpenCL.
I am having heterogeneous system with -
1) CPU
2) GPU
PFB my code base which is running in my system :
convolution.cl
// TODO: Add OpenCL kernel code here.
__kernel
void convolve(
const __global uint * const input,
__constant uint * const mask,
__global uint * const output,
const int inputWidth,
const int maskWidth){
const int x = get_global_id(0);
const int y = get_global_id(1);
uint sum = 0;
for (int r = 0; r < maskWidth; r++)
{
const int idxIntmp = (y + r) * inputWidth + x;
for (int c = 0; c < maskWidth; c++)
{
sum += mask[(r * maskWidth) + c] * input[idxIntmp + c];
}
}
output[y * get_global_size(0) + x] = sum;
}
and convolution.cpp -
//Convolution-Process of applying a 3×3 mask to an 8×8 input signal,resulting in a 6×6 output signal
#include "CL/cl.h"
#include "vector"
#include "iostream"
#include "time.h"
#include <fstream>
#include <sstream>
#include <string>
using namespace std;
// Constants
const unsigned int inputSignalWidth = 8;
const unsigned int inputSignalHeight = 8;
cl_uint inputSignal[inputSignalWidth][inputSignalHeight] =
{
{3, 1, 1, 4, 8, 2, 1, 3},
{4, 2, 1, 1, 2, 1, 2, 3},
{4, 4, 4, 4, 3, 2, 2, 2},
{9, 8, 3, 8, 9, 0, 0, 0},
{9, 3, 3, 9, 0, 0, 0, 0},
{0, 9, 0, 8, 0, 0, 0, 0},
{3, 0, 8, 8, 9, 4, 4, 4},
{5, 9, 8, 1, 8, 1, 1, 1}
};
const unsigned int outputSignalWidth = 6;
const unsigned int outputSignalHeight = 6;
cl_uint outputSignal[outputSignalWidth][outputSignalHeight];
const unsigned int maskWidth = 3;
const unsigned int maskHeight = 3;
cl_uint mask[maskWidth][maskHeight] =
{
{1, 1, 1},
{1, 0, 1},
{1, 1, 1},
};
inline void checkErr(cl_int err, const char * name)
{
if (err != CL_SUCCESS)
{
std::cerr << "ERROR: " << name
<< " (" << err << ")" << std::endl;
exit(EXIT_FAILURE);
}
}
void CL_CALLBACK contextCallback(
const char * errInfo,
const void * private_info,
size_t cb,
void * user_data)
{
std::cout << "Error occurred during context use: "<< errInfo << std::endl;
exit(EXIT_FAILURE);
}
int main(int argc,char argv[]){
cl_int errNum;
cl_uint numPlatforms;
cl_uint numDevices;
cl_platform_id * platformIDs;
cl_device_id * deviceIDs;
cl_context context = NULL;
cl_command_queue queue;
cl_program program;
cl_kernel kernel;
cl_mem inputSignalBuffer;
cl_mem outputSignalBuffer;
cl_mem maskBuffer;
double start,end,Totaltime;//Timer variables
errNum = clGetPlatformIDs(0, NULL, &numPlatforms);
checkErr(
(errNum != CL_SUCCESS) ? errNum :
(numPlatforms <= 0 ? -1 : CL_SUCCESS),
"clGetPlatformIDs");
platformIDs = (cl_platform_id *)malloc(sizeof(cl_platform_id) * numPlatforms);
errNum = clGetPlatformIDs(numPlatforms, platformIDs, NULL);
checkErr(
(errNum != CL_SUCCESS) ? errNum :
(numPlatforms <= 0 ? -1 : CL_SUCCESS), "clGetPlatformIDs");
deviceIDs = NULL;
cl_uint i;
for (i = 0; i < numPlatforms; i++)
{
errNum = clGetDeviceIDs(
platformIDs[i],
CL_DEVICE_TYPE_GPU,
0,
NULL,
&numDevices);
if (errNum != CL_SUCCESS && errNum != CL_DEVICE_NOT_FOUND)
{
checkErr(errNum, "clGetDeviceIDs");
}
else if (numDevices > 0)
{
deviceIDs = (cl_device_id *)malloc(
sizeof(cl_device_id) * numDevices);
errNum = clGetDeviceIDs(
platformIDs[i],
CL_DEVICE_TYPE_GPU,
numDevices,
&deviceIDs[0],
NULL);
checkErr(errNum, "clGetDeviceIDs");
break;
}
}
if (deviceIDs == NULL) {
std::cout << "No CPU device found" << std::endl;
exit(-1);
}
cl_context_properties contextProperties[] =
{
CL_CONTEXT_PLATFORM,(cl_context_properties)platformIDs[i], 0
};
context = clCreateContext(
contextProperties, numDevices, deviceIDs,
&contextCallback, NULL, &errNum);
checkErr(errNum, "clCreateContext");
std::ifstream srcFile("convolution.cl");
checkErr(srcFile.is_open() ? CL_SUCCESS : -1,
"reading convolution.cl");
std::string srcProg(
std::istreambuf_iterator<char>(srcFile),
(std::istreambuf_iterator<char>()));
const char * src = srcProg.c_str();
size_t length = srcProg.length();
program = clCreateProgramWithSource(context, 1, &src, &length, &errNum);
checkErr(errNum, "clCreateProgramWithSource");
errNum = clBuildProgram(program, numDevices, deviceIDs, NULL, NULL, NULL);
checkErr(errNum, "clBuildProgram");
kernel = clCreateKernel(program, "convolve", &errNum);
checkErr(errNum, "clCreateKernel");
inputSignalBuffer = clCreateBuffer(
context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
sizeof(cl_uint) * inputSignalHeight * inputSignalWidth,
static_cast<void *>(inputSignal), &errNum);
checkErr(errNum, "clCreateBuffer(inputSignal)");
maskBuffer = clCreateBuffer(
context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
sizeof(cl_uint) * maskHeight * maskWidth,
static_cast<void *>(mask), &errNum);
checkErr(errNum, "clCreateBuffer(mask)");
outputSignalBuffer = clCreateBuffer(
context, CL_MEM_WRITE_ONLY,
sizeof(cl_uint) * outputSignalHeight * outputSignalWidth,
NULL, &errNum);
checkErr(errNum, "clCreateBuffer(outputSignal)");
queue = clCreateCommandQueue(
context, deviceIDs[0], 0, &errNum);
checkErr(errNum, "clCreateCommandQueue");
errNum = clSetKernelArg(
kernel, 0, sizeof(cl_mem), &inputSignalBuffer);
errNum |= clSetKernelArg(
kernel, 1, sizeof(cl_mem), &maskBuffer);
errNum |= clSetKernelArg(
kernel, 2, sizeof(cl_mem), &outputSignalBuffer);
errNum |= clSetKernelArg(
kernel, 3, sizeof(cl_uint), &inputSignalWidth);
errNum |= clSetKernelArg(
kernel, 4, sizeof(cl_uint), &maskWidth);
checkErr(errNum, "clSetKernelArg");
const size_t globalWorkSize[1] ={ outputSignalWidth * outputSignalHeight };
const size_t localWorkSize[1] = { 1 };
start = clock();
errNum = clEnqueueNDRangeKernel(
queue,
kernel,
1,
NULL,
globalWorkSize,
localWorkSize,
0,
NULL,
NULL
);
checkErr(errNum, "clEnqueueNDRangeKernel");
errNum = clEnqueueReadBuffer(
queue, outputSignalBuffer, CL_TRUE, 0,
sizeof(cl_uint) * outputSignalHeight * outputSignalHeight,
outputSignal, 0, NULL, NULL);
checkErr(errNum, "clEnqueueReadBuffer");
end= clock(); - start;
cout<<"Time in ms = "<<((end/CLOCKS_PER_SEC) * 1000) << endl;
for (int y = 0; y < outputSignalHeight; y++)
{
for (int x = 0; x < outputSignalWidth; x++)
{
std::cout << outputSignal[x][y] << " ";
}
std::cout << std::endl;
}
return 0;
}
Questions :
I am having below doubts-
1) When I am using device type as CL_DEVICE_TYPE_GPU,
am getting 267 ms performance .When I am using CL_DEVICE_TYPE_CPU,execution time changed to 467 ms.
I want to know that what is the difference between running a convolution code on a CPU without GPU and CPU with GPU (by selecting device type as CL_DEVICE_TYPE_CPU) .
2) As I can see the convolution.cl file where there is a for loop which is executing 3 times.Can I call other Kernel for doing this operation from available kernel file ??
I am asking this question as I am new to the OpenCL coding and want to know that thing.

Both CPU & GPU are OpenCL Devices. So, by choosing CL_DEVICE_TYPE_CPU, you are telling OpenCL runtime to compile kernel code to CPU assembler & run it on CPU. When you are choosing CL_DEVICE_TYPE_GPU, kernel code is compiled to GPU assembler & executed on your video card. Ability to change device type without re-writing source code is of the main OpenCL features. It doesn't matter, does your CPU have integrated GPU, and / or discrete GPU is installed, you just pick available Device & run kernel on it.
For OpenCL 1.2 & older you can't call kernel from kernel. Dynamic parallelism is implemented in OpenCL 2.0.

For the first question: you should vectorize the kernel so opencl can easily use SIMD feature of your CPU hence unlock 4x(or 8x) more compute units per core.
__kernel
void convolve(
const __global uint8 * const input, // uint8 fits AVX(AVX2?) and uint4 fits SSE(SSE3?)
__constant uint8 * const mask,
__global uint8 * const output,
const int inputWidth,
const int maskWidth){
const int x = get_global_id(0); // this is 1/8 size now
const int y = get_global_id(1); // this is 1/8 size now
uint8 sum = 0; // a vector of 8 unsigneds
for (int r = 0; r < maskWidth; r++)
{
const int idxIntmp = (y + r) * inputWidth + x;
for (int c = 0; c < maskWidth; c++)
{
sum += mask[(r * maskWidth) + c] * input[idxIntmp + c]; //8 issued per clock
// scalars get promoted when used in direct multiplication of addition.
}
}
output[y * get_global_size(0) + x] = sum;
}
dont forget to decrease total work threads by 7/8 ratio (example: from 8k threads to 1k threads).
Please increase work per thread such as 50 convolutions per thread to increase occupation ratio of work units, then work on some local memory optimizations(for GPU) to get even better results such as 5ms per kernel..
On my AVX capable CPU, a simple matrix multiplication got speed up ratio of 2.4X going for 8-element vectorizations like this.
Running a kernel 3 times is not an issue if you offload enough work on it. If not, you should concatenate multiple kernels into a single one using some tricky algorithm.
If a profiler is not available at the moment, you can check GPU/CPU temperatures to get some idea of how close you are to the limits of hardware.
Play with number of local threads per work group. This can change performance as it lets more or less registers to be used per thread.

Related

OpenCL brute force TEA block 32bit, key 64bit

I decided to study OpenCL myself and write a brute-force password for the TEA algorithm, did I understand OpenCL correctly? can you improve something in the direction of speed? what mistakes have I made?
I prepare the first 5 bytes in cycles, the remaining 3 bytes are sorted out by the kernel, 255 threads at 65535 each
in the main program:
for (int x5 = KEY[0]; x5 >= 0; x5--) {
KEY[0]=x5;
for (int x4 = KEY[1]; x4 >= 0; x4--) {
KEY[1]=x4;
for (int x3 = KEY[2]; x3 >= 0; x3--) {
KEY[2]=x3;
for (int x2 = KEY[3]; x2 >= 0; x2--) {
KEY[3]=x2;
for (int x = KEY[4]; x >= 0; x--) {
KEY[4]=x;
ret = clEnqueueWriteBuffer(command_queue, key_mem_obj, CL_TRUE, 0,
8 * sizeof(int), KEY, 0, NULL, NULL);
ret = clEnqueueWriteBuffer(command_queue, cadr_mem_obj, CL_TRUE, 0,
1 * sizeof(int), CADR, 0, NULL, NULL);
ret = clSetKernelArg(kernel, 0, sizeof(cl_mem), (void *)&key_mem_obj);
ret = clSetKernelArg(kernel, 2, sizeof(cl_mem), (void *)&cadr_mem_obj);
NDRange = 0x0100;
ret = clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL,
&NDRange, NULL, 0, NULL, NULL);
if (ret != CL_SUCCESS) {
break;
}
ret = clEnqueueReadBuffer(command_queue, cadr_mem_obj, CL_TRUE, 0,
1 * sizeof(int), CADR, 0, NULL, NULL);
if (CADR[0]>0) {
uint16_t k=CADR[0];
ret = clEnqueueReadBuffer(command_queue, retc_mem_obj, CL_TRUE, 0,
524280 * sizeof(int), RETC, 0, NULL, NULL);
for ((i = 0); i < k; i++) {
Form1->Memo1->Lines->BeginUpdate();
Form1->Memo1->Lines->Add(IntToHex(RETC[i*8],2)+IntToHex(RETC[i*8+1],2)+
IntToHex(RETC[i*8+2],2)+IntToHex(RETC[i*8+3],2)+IntToHex(RETC[i*8+4],2)+
IntToHex(RETC[i*8+5],2)+IntToHex(RETC[i*8+6],2)+IntToHex(RETC[i*8+7],2));
Form1->Memo1->Lines->EndUpdate();
Form1->Label6->Caption=IntToStr(Form1->Memo1->Lines->Count-1);
}
CADR[0]=0;
}
KEY2[0]=KEY[0];
KEY2[1]=KEY[1];
KEY2[2]=KEY[2];
KEY2[3]=KEY[3];
KEY2[4]=KEY[4];
KEY2[5]=KEY[5];
KEY2[6]=KEY[6];
KEY2[7]=KEY[7];
if(Terminated){
break;
}
}
KEY[4]=0xFF;
}
KEY[3]=0xFF;
}
KEY[2]=0xFF;
}
KEY[1]=0xFF;
}
KEY[0]=0xFF;`
Kernel:
#pragma OPENCL EXTENSION cl_khr_byte_addressable_store : enable
#pragma OPENCL EXTENSION cl_khr_global_int32_base_atomics : enable
#pragma OPENCL EXTENSION cl_khr_local_int32_base_atomics : enable
#pragma OPENCL EXTENSION cl_khr_global_int32_extended_atomics : enable
#pragma OPENCL EXTENSION cl_khr_local_int32_extended_atomics : enable
__kernel void brute(__global const int *KEY, __global const int *DAT, __global int
*CADR,__global int *RETC)
{
int i = get_global_id(0);
ushort Data[2];
ushort Key[4];
Key[0]=(KEY[0]<<8)+KEY[1];
Key[1]=(KEY[2]<<8)+KEY[3];
// Key[2]=(KEY[4]<<8)+KEY[5];
Key[3]=(KEY[6]<<8)+KEY[7];
Key[2] = (KEY[4]<<8) + i;
for (int j=0xFFFF; j>=0; j--){
Key[3]=j;
Data[0]=(DAT[0]<<8)+DAT[1];
Data[1]=(DAT[2]<<8)+DAT[3];
ushort delta = 0x9e37;
ushort sum = (delta<<5);
for (uint n = 0;n < 32; ++n){
Data[1]-=(((Data[0])+Key[2])^(Data[0]+sum)^((Data[0]>>5)+Key[3]));
Data[0]-=(((Data[1]<<4)+Key[0])^(Data[1]+sum)^(Data[1]+Key[1]));
sum -= delta;
}
if ((Data[0]==0x0000) && (Data[1]==0x0000)){
int a=CADR[0];
atomic_inc(CADR);
RETC[a*8]=(Key[0] >> 8);
RETC[a*8+1]=(Key[0] & 0xFF);
RETC[a*8+2]=(Key[1] >> 8);
RETC[a*8+3]=(Key[1] & 0xFF);
RETC[a*8+4]=(Key[2] >> 8);
RETC[a*8+5]=(Key[2] & 0xFF);
RETC[a*8+6]=(Key[3] >> 8);
RETC[a*8+7]=(Key[3] & 0xFF);
}
}
}
If you only launch 256 threads that each do 65536 iterations of the same thing, your GPU will not be saturated and performance will be very poor. GPUs have thousands of "cores", and if you launch 256 threads you will only use 256 of them while the rest remains idle.
The idea of GPU parallelization is to split the work up into as many imdependent problems as there are. In your case this means: Lauch 256*65536 threads that do one Iteration each. Then performance will be much better.

OpenCL program works only for the multiple of itemsize

I'm new to openCL program and this is the problem I'm facing while executing a simple vector addition.
I have the following kernel code
#include <CL/cl.hpp>
#include<iostream>
#include <stdio.h>
#include <stdlib.h>
#define MAX_SOURCE_SIZE (0x100000)
int main() {
__kernel void vector_add(__global const int *A, __global const int *B, __global int *C) {
int i = get_global_id(0);
C[i] = A[i] + B[i];
}
I have integrated gpu and amd gpus on my system. I'm trying to perform vector addition on my intel gpu and for which I have installed the intel opencl drivers (i7 3rd gen processor with hd graphics).
I have the below openCL code
std::vector<cl::Platform> platforms;
cl::Platform::get(&platforms);
std::cout << "Total platforms including cpu: " << platforms.size() << std::endl;
if (platforms.size() == 0) {
std::cout << " No platforms found. Check OpenCL installation!\n";
exit(1);
}
int i;
const int LIST_SIZE = 50;
int *A = (int*)malloc(sizeof(int)*LIST_SIZE);
int *B = (int*)malloc(sizeof(int)*LIST_SIZE);
for(i = 0; i < LIST_SIZE; i++) {
A[i] = i;
B[i] = LIST_SIZE - i;
}
FILE *fp;
char *source_str;
size_t source_size;
fp = fopen("vector_add_kernel.cl", "r");
if (!fp) {
fprintf(stderr, "Failed to load kernel.\n");
exit(1);
}
source_str = (char*)malloc(MAX_SOURCE_SIZE);
source_size = fread( source_str, 1, MAX_SOURCE_SIZE, fp);
fclose( fp );
//std::cout<<source_str<<std::endl;
// Get platform and device information
cl_platform_id* platforms1 = NULL;
cl_device_id device_id = NULL;
cl_uint ret_num_devices;
cl_uint ret_num_platforms;
cl_int ret = clGetPlatformIDs(1, platforms1, &ret_num_platforms);
platforms1= (cl_platform_id*) malloc(sizeof(cl_platform_id) * ret_num_platforms);
clGetPlatformIDs(ret_num_platforms, platforms1, NULL);
/*
* Platform 0: Intel Graphics
* Platform 1 : AMD Graphics
*/
//CHANGE THE PLATFORM ACCORDING TO YOUR SYSTEM!!!!
ret = clGetDeviceIDs( platforms1[0], CL_DEVICE_TYPE_GPU, 1,
&device_id, &ret_num_devices);
// Create an OpenCL context
cl_context context = clCreateContext( NULL, 1, &device_id, NULL, NULL, &ret);
// Create a command queue
cl_command_queue command_queue = clCreateCommandQueue(context, device_id, 0, &ret);
// Create memory buffers on the device for each vector
cl_mem a_mem_obj = clCreateBuffer(context, CL_MEM_READ_ONLY,
LIST_SIZE * sizeof(int), NULL, &ret);
cl_mem b_mem_obj = clCreateBuffer(context, CL_MEM_READ_ONLY,
LIST_SIZE * sizeof(int), NULL, &ret);
cl_mem c_mem_obj = clCreateBuffer(context, CL_MEM_WRITE_ONLY,
LIST_SIZE * sizeof(int), NULL, &ret);
// Copy the lists A and B to their respective memory buffers
ret = clEnqueueWriteBuffer(command_queue, a_mem_obj, CL_TRUE, 0,
LIST_SIZE * sizeof(int), A, 0, NULL, NULL);
ret = clEnqueueWriteBuffer(command_queue, b_mem_obj, CL_TRUE, 0,
LIST_SIZE * sizeof(int), B, 0, NULL, NULL);
// Create a program from the kernel source
cl_program program = clCreateProgramWithSource(context, 1,
(const char **)&source_str, (const size_t *)&source_size, &ret);
// Build the program
ret = clBuildProgram(program, 1, &device_id, NULL, NULL, NULL);
// Create the OpenCL kernel
cl_kernel kernel = clCreateKernel(program, "vector_add", &ret);
// Set the arguments of the kernel
ret = clSetKernelArg(kernel, 0, sizeof(cl_mem), (void *)&a_mem_obj);
ret = clSetKernelArg(kernel, 1, sizeof(cl_mem), (void *)&b_mem_obj);
ret = clSetKernelArg(kernel, 2, sizeof(cl_mem), (void *)&c_mem_obj);
// Execute the OpenCL kernel on the list
size_t global_item_size = LIST_SIZE; // Process the entire lists
size_t local_item_size = 16; // Divide work items into groups of 64
ret = clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL,
&global_item_size, &local_item_size, 0, NULL, NULL);
// Read the memory buffer C on the device to the local variable C
int *C = (int*)malloc(sizeof(int)*LIST_SIZE);
ret = clEnqueueReadBuffer(command_queue, c_mem_obj, CL_TRUE, 0,
LIST_SIZE * sizeof(int), C, 0, NULL, NULL);
// Display the result to the screen
for(i = 0; i < LIST_SIZE; i++)
printf("%d + %d = %d\n", A[i], B[i], C[i]);
//FREE
return 0;
}
If the LISTSIZE is 50, it prints only till 48 that is 16*3. It prints only the multiple of LISTSIZE and I'm not able to figure out why?.
OpenCL kernels execute only for a multiple of the local thread block size (local Range, in your code local_item_size), which should not be smaller than 32 and must be a multiple of 2, (so it can be (32, 64, 128, 256, ...). If you set it to 16, half of the GPU will be idle at any time. global_item_size must be a multiple of local_item_size. You need at least 32 data items for the kernel to function and a lot more for it to yield good performance.
Also the part
#include <CL/cl.hpp>
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define MAX_SOURCE_SIZE (0x100000)
int main() {
is not OpenCL C code and does not belong in the .cl source file. If it is not too lengthy, you can write the OpenCL C code directly in the .cpp file as a raw string:
const string kernel_code = R"(
__kernel void vector_add(__global const int *A, __global const int *B, __global int *C) {
int i = get_global_id(0);
C[i] = A[i] + B[i];
}
)";
char* source_str = kernel_code.c_str();

What is the best practice to do reduce in OpenCL?

Imagine a binary operation (lets name it "+") with associative property. When you can compute a1 + a2 + a3 + a4 + ... in parallel, first computing
b1 = a1 + a2
b2 = a3 + a4
then
c1 = b1 + b2
c2 = b3 + b4
then doing the same thing for results of previous step, and so on, until there is one element left.
I'am learning OpenCL and trying to implement this approach to summarize all elements in array. I am a total newbie in this technology, so the program might look something weird.
This is the kernel:
__kernel void reduce (__global float *input, __global float *output)
{
size_t gl = get_global_id (0);
size_t s = get_local_size (0);
int i;
float accum = 0;
for (i=0; i<s; i++) {
accum += input[s*gl+i];
}
output[gl] = accum;
}
This is the main program:
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <CL/cl.h>
#define N (64*64*64*64)
#include <sys/time.h>
#include <stdlib.h>
double gettime ()
{
struct timeval tv;
gettimeofday (&tv, NULL);
return (double)tv.tv_sec + (0.000001 * (double)tv.tv_usec);
}
int main()
{
int i, fd, res = 0;
void* kernel_source = MAP_FAILED;
cl_context context;
cl_context_properties properties[3];
cl_kernel kernel;
cl_command_queue command_queue;
cl_program program;
cl_int err;
cl_uint num_of_platforms=0;
cl_platform_id platform_id;
cl_device_id device_id;
cl_uint num_of_devices=0;
cl_mem input, output;
size_t global, local;
cl_float *array = malloc (sizeof (cl_float)*N);
cl_float *array2 = malloc (sizeof (cl_float)*N);
for (i=0; i<N; i++) array[i] = i;
fd = open ("kernel.cl", O_RDONLY);
if (fd == -1) {
perror ("Cannot open kernel");
res = 1;
goto cleanup;
}
struct stat s;
res = fstat (fd, &s);
if (res == -1) {
perror ("Cannot stat() kernel");
res = 1;
goto cleanup;
}
kernel_source = mmap (NULL, s.st_size, PROT_READ, MAP_PRIVATE, fd, 0);
if (kernel_source == MAP_FAILED) {
perror ("Cannot map() kernel");
res = 1;
goto cleanup;
}
if (clGetPlatformIDs (1, &platform_id, &num_of_platforms) != CL_SUCCESS) {
printf("Unable to get platform_id\n");
res = 1;
goto cleanup;
}
if (clGetDeviceIDs(platform_id, CL_DEVICE_TYPE_GPU, 1, &device_id,
&num_of_devices) != CL_SUCCESS)
{
printf("Unable to get device_id\n");
res = 1;
goto cleanup;
}
properties[0]= CL_CONTEXT_PLATFORM;
properties[1]= (cl_context_properties) platform_id;
properties[2]= 0;
context = clCreateContext(properties,1,&device_id,NULL,NULL,&err);
command_queue = clCreateCommandQueue(context, device_id, 0, &err);
program = clCreateProgramWithSource(context, 1, (const char**)&kernel_source, NULL, &err);
if (clBuildProgram(program, 0, NULL, NULL, NULL, NULL) != CL_SUCCESS) {
char buffer[4096];
size_t len;
printf("Error building program\n");
clGetProgramBuildInfo (program, device_id, CL_PROGRAM_BUILD_LOG, sizeof (buffer), buffer, &len);
printf ("%s\n", buffer);
res = 1;
goto cleanup;
}
kernel = clCreateKernel(program, "reduce", &err);
if (err != CL_SUCCESS) {
printf("Unable to create kernel\n");
res = 1;
goto cleanup;
}
// create buffers for the input and ouput
input = clCreateBuffer(context, CL_MEM_READ_ONLY,
sizeof(cl_float) * N, NULL, NULL);
output = clCreateBuffer(context, CL_MEM_WRITE_ONLY,
sizeof(cl_float) * N, NULL, NULL);
// load data into the input buffer
clEnqueueWriteBuffer(command_queue, input, CL_TRUE, 0,
sizeof(cl_float) * N, array, 0, NULL, NULL);
size_t size = N;
cl_mem tmp;
double time = gettime();
while (size > 1)
{
// set the argument list for the kernel command
clSetKernelArg(kernel, 0, sizeof(cl_mem), &input);
clSetKernelArg(kernel, 1, sizeof(cl_mem), &output);
global = size;
local = 64;
// enqueue the kernel command for execution
clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global,
&local, 0, NULL, NULL);
clFinish(command_queue);
size = size/64;
tmp = output;
output = input;
input = tmp;
}
cl_float answer[1];
clEnqueueReadBuffer(command_queue, tmp, CL_TRUE, 0,
sizeof(cl_float), array, 0, NULL, NULL);
time = gettime() - time;
printf ("%f %f\n", array[0], time);
cleanup:
free (array);
free (array2);
clReleaseMemObject(input);
clReleaseMemObject(output);
clReleaseProgram(program);
clReleaseKernel(kernel);
clReleaseCommandQueue(command_queue);
clReleaseContext(context);
if (kernel_source != MAP_FAILED) munmap (kernel_source, s.st_size);
if (fd != -1) close (fd);
_Exit (res); // Kludge
return res;
}
So I re-run kernel until there is only one element in the buffer. Is this correct approach to compute sum of elements in OpenCL? The time which I measure with gettime is about 10 times slower when execution time of a simple loop on CPU (compiled clang 4.0.0 and -O2 -ffast-math flags). Hardware I use: Amd Ryzen 5 1600X and Amd Radeon HD 6950.
There's a couple of things you can do to try to improve performance.
Firstly, get rid of the clFinish call inside your loop. This forces individual executions of the kernels to be dependent on the entire state of the Command Queue reaching a synchronization point with the Host before continuing, which is unnecessary. The only synchronization required is that the kernels execute in order, and even if you have an out-of-order queue (which your program isn't requesting anyways), you can guarantee that with simple use of event objects.
size_t size = N;
size_t total_expected_events = 0;
for(size_t event_count = size; event_count > 1; event_count /= 64)
total_expected_events++;
cl_event * events = malloc(total_expected_events * sizeof(cl_event));
cl_mem tmp;
double time = gettime();
size_t event_index = 0;
while (size > 1)
{
// set the argument list for the kernel command
clSetKernelArg(kernel, 0, sizeof(cl_mem), &input);
clSetKernelArg(kernel, 1, sizeof(cl_mem), &output);
global = size;
local = 64;
if(event_index == 0)
// enqueue the kernel command for execution
clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global,
&local, 0, NULL, events);
else
clEnqueueNDRangeKernel(command_queue, kernel, 1, NULL, &global,
&local, 1, events + (event_index - 1), events + event_index);
size = size/64;
tmp = output;
output = input;
input = tmp;
event_index++;
}
clFinish(command_queue);
for(; event_index > 0; event_index--)
clReleaseEvent(events[event_index-1]);
free(events);
cl_float answer[1];
clEnqueueReadBuffer(command_queue, tmp, CL_TRUE, 0,
sizeof(cl_float), array, 0, NULL, NULL);
The other thing to potentially look into is performing the reduction all in one kernel, instead of spreading it out over multiple invocations of the same kernel. This is one potential example, though it may be more complicated than you need it to be.

OpenCL GPU calculation wrong

I am starting out OpenCL by converting existing C codes to an OpenCL. I am getting strange results with the both CPU and GPU calculation. Their values change 'every time' when I run the code. When I compare with the normal C, I would get 'somewhat' acceptable results from the CPU (but, still the results are not identical with the that of native C or even other languages), but when I run the 'exact same' code with GPU, I get gibberish results.
Here is my code on the Host
#include <stdio.h>
#include <stdlib.h>
#include <CL/cl.h>
#include <math.h>
double *arange(double start, double end, double step)
{
// 'arange' routine.
int i;
int arr_size = ((end - start) / step) + 1;
double *output = malloc(arr_size * sizeof(double));
for(i=0;i<arr_size;i++)
{
output[i] = start + (step * i);
}
return output;
}
int main()
{
// This code executes on the OpenCL Host
// Host data
double nu_ini = 100.0, nu_end = 2000.0, nu_step = 1.0;
double *delnu = arange(nu_ini, nu_end, nu_step);
double *nu, *inten, A, *gam_air, gam_self, E_pprime, *n_air, *del_air;
double *gamma, *f;
double prs = 950.0;
int i, j, dum, lines=0, ID, delnu_size = (((nu_end - nu_ini)/nu_step) + 1);
FILE *fp = fopen("h2o_HITRAN.par","r");
char string[320];
while(!feof(fp))
{
dum = fgetc(fp);
if(dum == '\n')
{
lines++;
}
}
rewind(fp);
nu = malloc(lines * sizeof(double));
inten = malloc(lines * sizeof(double));
gam_air = malloc(lines * sizeof(double));
n_air = malloc(lines * sizeof(double));
del_air = malloc(lines * sizeof(double));
gamma = malloc(lines * sizeof(double));
f = malloc(delnu_size * sizeof(double));
i=0;
while(fgets(string, 320, fp))
{
sscanf(string, "%2d %12lf %10le %10le %5lf %5lf %10lf %4lf %8lf", &ID, &nu[i], &inten[i], &A, &gam_air[i], &gam_self, &E_pprime, &n_air[i], &del_air[i]);
i++;
}
size_t line_siz = sizeof(double) * lines;
size_t delnu_siz = sizeof(double) * delnu_size;
// gamma calculation
for(i=0;i<lines;i++)
{
gamma[i] = pow((296.0/300.0),n_air[i]) * (gam_air[i]*(prs/1013.0));
}
// Use this to check the output of each API call
cl_int status;
// Retrieve the number of Platforms
cl_uint numPlatforms = 0;
status = clGetPlatformIDs(0, NULL, &numPlatforms);
// Allocate enough space for each Platform
cl_platform_id *platforms = NULL;
platforms = (cl_platform_id*)malloc(numPlatforms*sizeof(cl_platform_id));
// Fill in the Platforms
status = clGetPlatformIDs(numPlatforms, platforms, NULL);
// Retrieve the number of Devices
cl_uint numDevices = 0;
status = clGetDeviceIDs(platforms[0],CL_DEVICE_TYPE_ALL, 0, NULL, &numDevices);
// Allocate enough spaces for each Devices
char name_data[100];
int *comp_units;
cl_device_fp_config cfg;
cl_device_id *devices = NULL;
devices = (cl_device_id*)malloc(numDevices*sizeof(cl_device_id));
// Fill in the Devices
status = clGetDeviceIDs(platforms[0], CL_DEVICE_TYPE_ALL, numDevices, devices, NULL);
// Create a context and associate it with the devices
cl_context context = NULL;
context = clCreateContext(NULL, numDevices, devices, NULL, NULL, &status);
// Create a command queue and associate it with the devices
cl_command_queue cmdQueue = NULL;
cmdQueue = clCreateCommandQueueWithProperties(context, devices[0], 0, &status);
// Create a buffer objects that will contain the data from the host array 'buf_xxxx'
cl_mem buf_inten = NULL;
cl_mem buf_gamma = NULL;
cl_mem buf_delnu = NULL;
cl_mem buf_nu = NULL;
cl_mem buf_del_air = NULL;
cl_mem buf_f = NULL;
buf_inten = clCreateBuffer(context, CL_MEM_READ_ONLY, line_siz, NULL, &status);
buf_gamma = clCreateBuffer(context, CL_MEM_READ_ONLY, line_siz, NULL, &status);
buf_delnu = clCreateBuffer(context, CL_MEM_READ_ONLY, delnu_siz, NULL, &status);
buf_nu = clCreateBuffer(context, CL_MEM_READ_ONLY, line_siz, NULL, &status);
buf_del_air = clCreateBuffer(context, CL_MEM_READ_ONLY, line_siz, NULL, &status);
buf_f = clCreateBuffer(context, CL_MEM_READ_ONLY, delnu_siz, NULL, &status);
// Write input array A to the Device buffer 'buf_xxx'
status = clEnqueueWriteBuffer(cmdQueue, buf_inten, CL_FALSE, 0, line_siz, inten, 0, NULL, NULL);
status = clEnqueueWriteBuffer(cmdQueue, buf_gamma, CL_FALSE, 0, line_siz, gamma, 0, NULL, NULL);
status = clEnqueueWriteBuffer(cmdQueue, buf_delnu, CL_FALSE, 0, delnu_siz, delnu, 0, NULL, NULL);
status = clEnqueueWriteBuffer(cmdQueue, buf_nu, CL_FALSE, 0, line_siz, nu, 0, NULL, NULL);
status = clEnqueueWriteBuffer(cmdQueue, buf_del_air, CL_FALSE, 0, line_siz, del_air, 0, NULL, NULL);
// Create Program with the source code
cl_program program = NULL;
size_t program_size;
char *program_Source;
FILE *program_handle = fopen("abs_calc.cl","r");
fseek(program_handle, 0, SEEK_END);
program_size = ftell(program_handle);
rewind(program_handle);
program_Source = (char*)malloc(program_size+1);
program_Source[program_size] = '\0';
fread(program_Source, sizeof(char), program_size, program_handle);
fclose(program_handle);
program = clCreateProgramWithSource(context, 1, (const char**)&program_Source, &program_size, &status);
// Compile the Program for the Device
status = clBuildProgram(program, numDevices, devices, NULL, NULL, NULL);
// Create the vector addition kernel
cl_kernel kernel = NULL;
kernel = clCreateKernel(program, "abs_cross", &status);
// Associate the input and output buffers with the kernel
status = clSetKernelArg(kernel, 0, sizeof(cl_mem), &buf_inten);
status = clSetKernelArg(kernel, 1, sizeof(cl_mem), &buf_gamma);
status = clSetKernelArg(kernel, 2, sizeof(cl_mem), &buf_delnu);
status = clSetKernelArg(kernel, 3, sizeof(cl_mem), &buf_nu);
status = clSetKernelArg(kernel, 4, sizeof(cl_mem), &buf_del_air);
status = clSetKernelArg(kernel, 5, sizeof(cl_mem), &buf_f);
// Define index space (global work size) of work items for execution.
// A workgroup size (local work size) is not required, but can be used.
size_t globalWorkSize[2] = {lines, delnu_size};
// Execute the kernel for execution
status = clEnqueueNDRangeKernel(cmdQueue, kernel, 2, NULL, globalWorkSize, NULL, 0, NULL, NULL);
// Read the Device output buffer to the host output array
clEnqueueReadBuffer(cmdQueue, buf_f, CL_TRUE, 0, delnu_siz, f, 0, NULL, NULL);
// Verify the output
FILE *file = fopen("opencl_output","w");
for(i=0;i<delnu_size;i++)
{
fprintf(file, "%le %le\n", delnu[i], f[i]);
}
// Free OpenCL resources
clReleaseKernel(kernel);
clReleaseProgram(program);
clReleaseCommandQueue(cmdQueue);
clReleaseMemObject(buf_nu);
clReleaseMemObject(buf_inten);
clReleaseMemObject(buf_del_air);
clReleaseMemObject(buf_gamma);
clReleaseMemObject(buf_f);
clReleaseMemObject(buf_delnu);
clReleaseContext(context);
// Free host resources
free(nu);
free(inten);
free(gam_air);
free(n_air);
free(del_air);
free(delnu);
free(gamma);
free(f);
free(platforms);
free(devices);
fclose(fp);
fclose(file);
return 0;
}
and this is my kernel code
#pragma OPENCL EXTENSION cl_khr_fp64 : enable
kernel void abs_cross(global double *inten,
global double *gamma,
global double *delnu,
global double *nu,
global double *del_air,
global double *f)
{
double pie = 4.0*atan(1.0);
int i = get_global_id(0);
int j = get_global_id(1);
f[j] += inten[i] * ((1.0/pie) * (gamma[i] / (pown(gamma[i],2) + pown((delnu[j] - nu[i] + del_air[i] * 950.0/1013.0),2))));
}
Am I doing something wrong?
Thank you.
You appear to be running a 2D global work size, but storing into a location based only on dimension 1 (not 0). Therefore multiple work items are storing into the same location using +=. You have a race condition. You could use atomics to solve this, but it will likely slow the performance down too much. Therefore, you should store intermediate results and then do a parallel reduction operation.
I am using AMD W2100, and yes, I have printed out all the supported extension and it included cl_khr_fp64 extension.
Sorry, I forgot to include the original calculation. The actual calculation goes like the following..
for(i=0,i<lines;i++)
{
for(j=0;j<delnu_size;j++)
{
f[j] += inten[i] * ((1.0/pie) * (gamma[i] / (pow(gamma[i],2) + pow((delnu[j] - nu[i] + del_air[i] * 950.0/1013.0),2))));
}
}
I would write OpenCL kernel as below,
Without using atomics and only single work dimension.
global_work_size = delnu_size
There could be a better way but its the simplest one.
__kernel void test(__global double *gamma,
__global double *inten,
__global double *delnu,
__global double *delair,
__global double *f,
const int lines)
{
double pie = 4.0*atan(1.0);
int j = get_global_id(0);
f[j] = 0;
for(i=0,i<lines;i++)
{
f[j] += inten[i] * ((1.0/pie) * (gamma[i] / (pow(gamma[i],2) + pow((delnu[j] - nu[i] + del_air[i] * 950.0/1013.0),2))));
}
}
You need to understand how OpenCL kernel is executed.
You can think of it as large number of threads executing concurrently
and each thread could be identified with get_global_id

How to remove CL_INVALID_PLATFORM error in opencl code?

Doing simple matrix multiplication using OpenCL:
// Multiply two matrices A * B = C
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <oclUtils.h>
#define WA 3
#define HA 3
#define WB 3
#define HB 3
#define WC 3
#define HC 3
// Allocates a matrix with random float entries.
void randomInit(float* data, int size)
{
for (int i = 0; i < size; ++i)
data[i] = rand() / (float)RAND_MAX;
}
/////////////////////////////////////////////////////////
// Program main
/////////////////////////////////////////////////////////
int
main(int argc, char** argv)
{
// set seed for rand()
srand(2006);
// 1. allocate host memory for matrices A and B
unsigned int size_A = WA * HA;
unsigned int mem_size_A = sizeof(float) * size_A;
float* h_A = (float*) malloc(mem_size_A);
unsigned int size_B = WB * HB;
unsigned int mem_size_B = sizeof(float) * size_B;
float* h_B = (float*) malloc(mem_size_B);
// 2. initialize host memory
randomInit(h_A, size_A);
randomInit(h_B, size_B);
// 3. print out A and B
printf("\n\nMatrix A\n");
for(int i = 0; i < size_A; i++)
{
printf("%f ", h_A[i]);
if(((i + 1) % WA) == 0)
printf("\n");
}
printf("\n\nMatrix B\n");
for(int i = 0; i < size_B; i++)
{
printf("%f ", h_B[i]);
if(((i + 1) % WB) == 0)
printf("\n");
}
// 4. allocate host memory for the result C
unsigned int size_C = WC * HC;
unsigned int mem_size_C = sizeof(float) * size_C;
float* h_C = (float*) malloc(mem_size_C);
// 5. Initialize OpenCL
// OpenCL specific variables
cl_context clGPUContext;
cl_command_queue clCommandQue;
cl_program clProgram;
cl_kernel clKernel;
size_t dataBytes;
size_t kernelLength;
cl_int errcode;
// OpenCL device memory for matrices
cl_mem d_A;
cl_mem d_B;
cl_mem d_C;
/*****************************************/
/* Initialize OpenCL */
/*****************************************/
clGPUContext = clCreateContextFromType(0,
CL_DEVICE_TYPE_GPU,
NULL, NULL, &errcode);
shrCheckError(errcode, CL_SUCCESS);
// get the list of GPU devices associated
// with context
errcode = clGetContextInfo(clGPUContext,
CL_CONTEXT_DEVICES, 0, NULL,
&dataBytes);
cl_device_id *clDevices = (cl_device_id *)
malloc(dataBytes);
errcode |= clGetContextInfo(clGPUContext,
CL_CONTEXT_DEVICES, dataBytes,
clDevices, NULL);
//shrCheckError(errcode, CL_SUCCESS);
//Create a command-queue
clCommandQue = clCreateCommandQueue(clGPUContext,
clDevices[0], 0, &errcode);
//shrCheckError(errcode, CL_SUCCESS);
// Setup device memory
d_C = clCreateBuffer(clGPUContext,
CL_MEM_READ_WRITE,
mem_size_A, NULL, &errcode);
d_A = clCreateBuffer(clGPUContext,
CL_MEM_READ_WRITE | CL_MEM_COPY_HOST_PTR,
mem_size_A, h_A, &errcode);
d_B = clCreateBuffer(clGPUContext,
CL_MEM_READ_WRITE | CL_MEM_COPY_HOST_PTR,
mem_size_B, h_B, &errcode);
// 6. Load and build OpenCL kernel
char *clMatrixMul = oclLoadProgSource("kernel.cl",
"// My comment\n",
&kernelLength);
//shrCheckError(clMatrixMul != NULL, shrTRUE);
clProgram = clCreateProgramWithSource(clGPUContext,
1, (const char **)&clMatrixMul,
&kernelLength, &errcode);
//shrCheckError(errcode, CL_SUCCESS);
errcode = clBuildProgram(clProgram, 0,
NULL, NULL, NULL, NULL);
//shrCheckError(errcode, CL_SUCCESS);
clKernel = clCreateKernel(clProgram,
"matrixMul", &errcode);
//shrCheckError(errcode, CL_SUCCESS);
// 7. Launch OpenCL kernel
size_t localWorkSize[2], globalWorkSize[2];
int wA = WA;
int wC = WC;
errcode = clSetKernelArg(clKernel, 0,
sizeof(cl_mem), (void *)&d_C);
errcode |= clSetKernelArg(clKernel, 1,
sizeof(cl_mem), (void *)&d_A);
errcode |= clSetKernelArg(clKernel, 2,
sizeof(cl_mem), (void *)&d_B);
errcode |= clSetKernelArg(clKernel, 3,
sizeof(int), (void *)&wA);
errcode |= clSetKernelArg(clKernel, 4,
sizeof(int), (void *)&wC);
//shrCheckError(errcode, CL_SUCCESS);
localWorkSize[0] = 3;
localWorkSize[1] = 3;
globalWorkSize[0] = 3;
globalWorkSize[1] = 3;
errcode = clEnqueueNDRangeKernel(clCommandQue,
clKernel, 2, NULL, globalWorkSize,
localWorkSize, 0, NULL, NULL);
//shrCheckError(errcode, CL_SUCCESS);
// 8. Retrieve result from device
errcode = clEnqueueReadBuffer(clCommandQue,
d_C, CL_TRUE, 0, mem_size_C,
h_C, 0, NULL, NULL);
//shrCheckError(errcode, CL_SUCCESS);
// 9. print out the results
printf("\n\nMatrix C (Results)\n");
for(int i = 0; i < size_C; i++)
{
printf("%f ", h_C[i]);
if(((i + 1) % WC) == 0)
printf("\n");
}
printf("\n");
// 10. clean up memory
free(h_A);
free(h_B);
free(h_C);
clReleaseMemObject(d_A);
clReleaseMemObject(d_C);
clReleaseMemObject(d_B);
free(clDevices);
free(clMatrixMul);
clReleaseContext(clGPUContext);
clReleaseKernel(clKernel);
clReleaseProgram(clProgram);
clReleaseCommandQueue(clCommandQue);
}
In the above code I keep getting error at the place :
/**********************/ / Initialize OpenCL
/ /**********************/
clGPUContext = clCreateContextFromType(0,
CL_DEVICE_TYPE_GPU,
NULL, NULL, &errcode); shrCheckError(errcode, CL_SUCCESS);
The error code being returned is -32 that means: CL_INVALID_PLATFORM"
How do I remove this error?
OS: Windows 7, 32 bit, NVIDIA GPU GeForce 610
The Nvidia drivers expect you to provide a non-NULL properties pointer as first argument to the clCreateContextFromType call.
The Khronos specification for clCreateContextFromType states that if NULL is passed for the properties parameter, the platform that is selected is implementation dependent. In case of Nvidia the choice seems to be that no platform at all is selected if a NULL pointer is passed. See clCreateContextFromType for more information.
On the other hand, this behavior is consistent with Issue #3 in the cl_khr_icd extension, which would apply if you are using OpenCL through the ICD, and which states:
3: How will the ICD handle a NULL cl_platform_id?
RESOLVED: The NULL platform is not supported by the ICD.
To pass the properties to clCreateContextFromType, first query the platforms with clGetPlatformIDs. Then construct a properties array with the desired platform ID and pass it to clCreateContextFromType. Something along the following lines should work with a C99 compliant compiler:
// query the number of platforms
cl_uint numPlatforms;
errcode = clGetPlatformIDs(0, NULL, &numPlatforms);
shrCheckError(errcode, CL_SUCCESS);
// now get all the platform IDs
cl_platform_id platforms[numPlatforms];
errcode = clGetPlatformIDs(numPlatforms, platforms, NULL);
shrCheckError(errcode, CL_SUCCESS);
// set platform property - we just pick the first one
cl_context_properties properties[] = {CL_CONTEXT_PLATFORM, (int) platforms[0], 0};
clGPUContext = clCreateContextFromType(properties, CL_DEVICE_TYPE_GPU, NULL, NULL, &errcode);
shrCheckError(errcode, CL_SUCCESS);

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