clEnqueueReadBuffer return a void buffer - opencl

I am trying to implement an OpenCL project that should take in input two "grey" channels images and return a "grey" channel buffer
My OpenCL code:
__kernel void motion_detection(__global uchar* bgImage, __global uchar* motionImage,__global uchar* outputImage)
{
int i = get_global_id(0);
int tsh = 7;
uchar bgEl = bgImage[i];
uchar motionEl = motionImage[i];
float bg = convert_float(bgEl);
float mo = convert_float(motionEl);
float temp = fabs(bg-mo);
if(temp>tsh){temp=255.0;
}else
{temp=0.0;
}
outputImage[i] = convert_uchar(bg);
}
My Host:
// get width and height of input image
height = inputImage->rows;
width = inputImage->cols;
// get the pointer to pixel data
cl_uchar* inputImageData = inputImage->data;
cl_uchar* motionImageData = motionImage->data;
cl_uchar* outputImageData = outputImage->data;
After set up buffers:
// Create memory object for input Image
inputImageBuffer = clCreateBuffer(
context,
CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
width * height * greyPixelSize,
inputImageData,
&status);
CHECK_OPENCL_ERROR(status, "clCreateBuffer failed. (inputImageBuffer)");
motionImageBuffer = clCreateBuffer(
context,
CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR,
width * height * greyPixelSize,
motionImageData,
&status);
CHECK_OPENCL_ERROR(status, "clCreateBuffer failed. (inputImageBuffer)");
// Create memory objects for output Image
outputImageBuffer = clCreateBuffer(
context,
CL_MEM_READ_WRITE,
width * height * greyPixelSize,
0,
&status);
Settting up arguments and executing OpenCL kernel:
// input buffer image
status = clSetKernelArg(
kernl,
0,
sizeof(cl_mem),
&inputImageBuffer);
CHECK_OPENCL_ERROR(status, "clSetKernelArg failed. (inputImageBuffer)");
status = clSetKernelArg(
kernl,
1,
sizeof(cl_mem),
&motionImageBuffer);
CHECK_OPENCL_ERROR(status, "clSetKernelArg failed. (inputImageBuffer)");
// outBuffer imager
status = clSetKernelArg(
kernl,
2,
sizeof(cl_mem),
&outputImageBuffer);
CHECK_OPENCL_ERROR(status, "clSetKernelArg failed. (outputImageBuffer)");
status = clEnqueueNDRangeKernel(
commandQueue,
kernl,
1,
NULL,
globalThreads,
localThreads,
0,
NULL,
&ndrEvt);
clFinish(commandQueue);
Here is my problem, when this intruction is executed it returns outputImageData as "\0" and nothing else:
status = clEnqueueReadBuffer(
commandQueue,
outputImageBuffer,
CL_TRUE,
0,
width * height * greyPixelSize,
outputImageData,
0,
NULL,
NULL);
CHECK_OPENCL_ERROR(status, "clEnqueueReadBuffer failed.");

Related

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

openCL Long Overflowing

Before I start I am a C beginner and I am trying to do some openCL work which might have been a mistake. Below is my kernel code:
__kernel void collatz(__global int* in, __global int* out)
{
uint id = get_global_id(0);
unsigned long n = (unsigned long)id;
uint count = 0;
while (n > 1) {
if (n % 2 == 0) {
n = n / 2;
} else {
if(n == 1572066143) {
unsigned long test = n;
printf("BEFORE - %lu\n", n);
test = (3 * test) + 1;
printf("AFTER - %lu\n", test);
n = (3 * n) + 1;
} else {
n = (3 * n) + 1;
}
}
count = count + 1;
}
out[id] = count;
}
and the output:
BEFORE - 1572066143
AFTER - 421231134
To me it looks like n is overflowing but I can't figure out why it is happening.
The interesting thing is if I create a new variable to store the same value as n then it seems to work correctly.
unsigned long test = 1572066143;
printf("BEFORE - %lu\n", test);
test = (3 * test) + 1;
printf("AFTER - %lu\n", test);
Output:
BEFORE - 1572066143
AFTER - 4716198430
As I said I am a C beginner so I could be doing something very stupid! Any help would be appreciated as I have been pulling my hair out for hours now!
Thanks,
Stephen
Update:
Here is my host code in case I am doing something stupid on that end:
int _tmain(int argc, _TCHAR* argv[])
{
/*Step1: Getting platforms and choose an available one.*/
cl_uint numPlatforms; //the NO. of platforms
cl_platform_id platform = NULL; //the chosen platform
cl_int status = clGetPlatformIDs(0, NULL, &numPlatforms);
cl_platform_id* platforms = (cl_platform_id*)malloc(numPlatforms* sizeof(cl_platform_id));
status = clGetPlatformIDs(numPlatforms, platforms, NULL);
platform = platforms[0];
free(platforms);
/*Step 2:Query the platform and choose the first GPU device if has one.*/
cl_device_id *devices;
devices = (cl_device_id*)malloc(1 * sizeof(cl_device_id));
clGetDeviceIDs(platform, CL_DEVICE_TYPE_GPU, 1, devices, NULL);
/*Step 3: Create context.*/
cl_context context = clCreateContext(NULL, 1, devices, NULL, NULL, NULL);
/*Step 4: Creating command queue associate with the context.*/
cl_command_queue commandQueue = clCreateCommandQueue(context, devices[0], 0, NULL);
/*Step 5: Create program object */
const char *filename = "HelloWorld_Kernel.cl";
std::string sourceStr;
status = convertToString(filename, sourceStr);
const char *source = sourceStr.c_str();
size_t sourceSize[] = { strlen(source) };
cl_program program = clCreateProgramWithSource(context, 1, &source, sourceSize, NULL);
status = clBuildProgram(program, 1, devices, NULL, NULL, NULL);
/*Step 7: Initial input,output for the host and create memory objects for the kernel*/
cl_ulong max = 2000000;
cl_ulong *numbers = NULL;
numbers = new cl_ulong[max];
for (int i = 1; i <= max; i++) {
numbers[i] = i;
}
int *output = (int*)malloc(sizeof(cl_ulong) * max);
cl_mem inputBuffer = clCreateBuffer(context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR, max * sizeof(cl_ulong), (void *)numbers, NULL);
cl_mem outputBuffer = clCreateBuffer(context, CL_MEM_WRITE_ONLY, max * sizeof(cl_ulong), NULL, NULL);
/*Step 8: Create kernel object */
cl_kernel kernel = clCreateKernel(program, "collatz", NULL);
/*Step 9: Sets Kernel arguments.*/
status = clSetKernelArg(kernel, 0, sizeof(cl_mem), (void *)&inputBuffer);
// Determine the size of the log
size_t log_size;
clGetProgramBuildInfo(program, devices[0], CL_PROGRAM_BUILD_LOG, 0, NULL, &log_size);
// Allocate memory for the log
char *log = (char *)malloc(log_size);
// Get the log
clGetProgramBuildInfo(program, devices[0], CL_PROGRAM_BUILD_LOG, log_size, log, NULL);
// Print the log
printf("%s\n", log);
status = clSetKernelArg(kernel, 1, sizeof(cl_mem), (void *)&outputBuffer);
/*Step 10: Running the kernel.*/
size_t global_work_size[] = { max };
status = clEnqueueNDRangeKernel(commandQueue, kernel, 1, NULL, global_work_size, NULL, 0, NULL, NULL);
/*Step 11: Read the data put back to host memory.*/
status = clEnqueueReadBuffer(commandQueue, outputBuffer, CL_TRUE, 0, max * sizeof(cl_ulong), output, 0, NULL, NULL);
return SUCCESS;
}
I finally got to the bottom of the issue.
I was running the code on my Intel HD Graphics 4600 chip and it was producing the strange behaviour shown in the original question. I switched to using my AMD card and then it started working as expected!
Very strange. Thanks to everyone for their help!
Host side and device size values have different sizes. In host, long can vary from 32 to 64bits, depending on the platform. In device, long refers to 64bits only.
printf() function, as defined in C says that %ld is to print long (host side long) numbers. You are using printf in a kernel, so.... It could be that the C-like parser is used, therefore printing the variable as a 32bits long.
Can you try printing it as %lld or as a floating point?

OpenCL: Can one kernel call the other kernel

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.

Transmission parameters in to the function in openCL

Have following kernel function:
private static String programSource =
"__kernel void sampleKernel(__global float *Y, __global float *param) "
+ "{ int index = get_global_id(0); "
+ " Y[index]=param[0]-Y[index]/param[1]-param[2]; "
+ "} ";
First argument "Y" works perfect, but second parameter "param" works non correct, I receive null values . Second parametr must be a array and consists from 3 cells.
Fragment of code with the transmission parameters:
float[] arr_params = new float[3];
arr_params[0] = (float) h_c;
arr_params[1] = (float) sy;
arr_params[2] = (float) dy;
//pointers
Pointer Pvy = Pointer.to(vy);
Pointer Parr_params = Pointer.to(arr_params);
cl_mem memObjects[] = new cl_mem[2];
memObjects[0] = clCreateBuffer(context,
CL_MEM_READ_WRITE,
Sizeof.cl_float * vy.length, Pvy, null);
memObjects[1] = clCreateBuffer(context,
CL_MEM_READ_ONLY,
Sizeof.cl_float * arr_params.length, Parr_params, null);
// Set the arguments for the kernel
clSetKernelArg(kernel, 0,
Sizeof.cl_mem, Pointer.to(memObjects[0]));
clSetKernelArg(kernel, 1,
Sizeof.cl_mem, Pointer.to(memObjects[1]));
// Set the work-item dimensions
long global_work_size[] = new long[]{vy.length};
long local_work_size[] = new long[]{1};
// Execute the kernel
clEnqueueNDRangeKernel(commandQueue, kernel, 1, null,
global_work_size, local_work_size, 0, null, null);
// Read the output data
clEnqueueReadBuffer(commandQueue, memObjects[0], CL_TRUE, 0, vy.length * Sizeof.cl_float, Pvy, 0, null, null);
// Release kernel, program, and memory objects
clReleaseMemObject(memObjects[0]);
clReleaseMemObject(memObjects[1]);
The second buffer is all zeros because, in the clCreateBuffer call, you haven't told OpenCL where to get the data. Use CL_MEM_USE_HOST_PTR or CL_MEM_COPY_HOST_PTR.

OpenCL local_work_size issues

I have a vector of size n (unknown before the main program is launched), and I want to experiment with the work-group sizes. However, unless I set the local_work_size to a number that exactly divides n, I only get 0 values in fprop below.
Kernel:
__kernel void palt(__global double *fprop, __global const double *fcoll,
__global const int *nn, const uint max_size)
{
size_t l = get_global_id(0);
if( l > max_size ) return;
fprop[l] = fcoll[nn[l]];
}
Host code:
int block_sz_p = 128;
const int max_size = ns*imax;
// set the parameters for the propagation operator
errNum = clSetKernelArg(propagation_kernel, 0, sizeof(cl_mem), &fpd);
errNum |= clSetKernelArg(propagation_kernel, 1, sizeof(cl_mem), &fcd);
errNum |= clSetKernelArg(propagation_kernel, 2, sizeof(cl_mem), &nnd);
errNum |= clSetKernelArg(propagation_kernel, 3, sizeof(int), (void *) &max_sz);
checkErr(errNum, "clSetKernelArg(propagation)");
// specify the work group size/dim
const size_t work_dim = 3;
const size_t global_work_size_propagation[] = {imax*ns, 1, 1};
const size_t local_work_size_propagation[] = {block_sz_p, 1, 1};
// propagation
clEnqueueNDRangeKernel(queue, propagation_kernel, work_dim, NULL,
global_work_size_propagation, local_work_size_propagation,
0, NULL, &event);
clWaitForEvents(1, &event);
clGetEventProfilingInfo(event, CL_PROFILING_COMMAND_START,
sizeof(cl_ulong), &start, NULL);
clGetEventProfilingInfo(event, CL_PROFILING_COMMAND_END,
sizeof(cl_ulong), &end, NULL);
tker2 = (end-start);
What's going on here?
You should check for CL errors, clEnqueueNDRangeKernel and other calls return a error code (others return the error code by reference).
I assume the problem is when the global workgroup size isn't divisible by the local workgroup size, which is not supported and generates a CL error:
CL_INVALID_WORK_GROUP_SIZE if local_work_size is specified and number of work-items specified by global_work_size is not evenly divisable by size of work-group given by local_work_size or does not match the work-group size specified for kernel using the attribute((reqd_work_group_size(X, Y, Z))) qualifier in program source.
From the clEnqueueNDRangeKernel man page.

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