I don’t know of any “special” requirements to have overlap of two kernels issued to two separate command queues. Here is a simple example:
$ cat t13.cpp
#include <CL/opencl.h>
#include <stdint.h>
#include <stdio.h>
#include <inttypes.h>
#include <stdlib.h>
const char source[] =
"__kernel void test_rotate(__global ulong *d_count, ulong loops, ulong patt)"
"{"
" ulong n = patt;"
" for (ulong i = 0; i<loops; i++)"
" n &= (107 << (patt+(i%7)));"
" d_count[0] = n + loops;"
"}"
;
int main(int argc, char *argv[])
{
cl_platform_id platform;
cl_device_id device;
cl_context context;
cl_command_queue queue1, queue2;
cl_program program;
cl_mem mem1, mem2;
cl_kernel kernel;
bool two_kernels = false;
unsigned long long loops = 1000;
if (argc > 1) loops *= atoi(argv[1]);
if (argc > 2) two_kernels = true;
if (two_kernels) printf("running two kernels\n");
else printf("running one kernel\n");
printf("running %lu loops\n", loops);
unsigned long long pattern = 1;
clGetPlatformIDs(1, &platform, NULL);
clGetDeviceIDs(platform, CL_DEVICE_TYPE_ALL, 1, &device, NULL);
context = clCreateContext(NULL, 1, &device, NULL, NULL, NULL);
queue1 = clCreateCommandQueue(context, device, CL_QUEUE_OUT_OF_ORDER_EXEC_MODE_ENABLE, NULL);
queue2 = clCreateCommandQueue(context, device, CL_QUEUE_OUT_OF_ORDER_EXEC_MODE_ENABLE, NULL);
const char *sources[1] = {source};
program = clCreateProgramWithSource(context, 1, sources, NULL, NULL);
clBuildProgram(program, 1, &device, NULL, NULL, NULL);
mem1 = clCreateBuffer(context, CL_MEM_READ_WRITE, 1*sizeof(cl_ulong), NULL, NULL);
mem2 = clCreateBuffer(context, CL_MEM_READ_WRITE, 1*sizeof(cl_ulong), NULL, NULL);
kernel = clCreateKernel(program, "test_rotate", NULL);
const size_t work_size[1] = {1};
clSetKernelArg(kernel, 0, sizeof(mem1), &mem1);
clSetKernelArg(kernel, 1, sizeof(loops), &loops);
clSetKernelArg(kernel, 2, sizeof(pattern), &pattern);
clEnqueueNDRangeKernel(queue1, kernel, 1, NULL, work_size, work_size, 0, NULL, NULL);
if (two_kernels){
clSetKernelArg(kernel, 0, sizeof(mem2), &mem2);
clSetKernelArg(kernel, 1, sizeof(loops), &loops);
clSetKernelArg(kernel, 2, sizeof(pattern), &pattern);
clEnqueueNDRangeKernel(queue2, kernel, 1, NULL, work_size, work_size, 0, NULL, NULL);
}
cl_ulong *buf1 = (cl_ulong *)clEnqueueMapBuffer(queue1, mem1, true, CL_MAP_READ, 0, 1*sizeof(cl_ulong), 0, NULL, NULL, NULL);
cl_ulong *buf2 = (cl_ulong *)clEnqueueMapBuffer(queue2, mem2, true, CL_MAP_READ, 0, 1*sizeof(cl_ulong), 0, NULL, NULL, NULL);
printf("result1: %lu\n", buf1[0]);
printf("result2: %lu\n", buf2[0]);
clEnqueueUnmapMemObject(queue1, mem1, buf1, 0, NULL, NULL);
clEnqueueUnmapMemObject(queue2, mem2, buf2, 0, NULL, NULL);
return 0;
}
$ g++ t13.cpp -o t13 -I/usr/local/cuda/include -lOpenCL
$ time ./t13
running one kernel
running 1000 loops
result1: 1000
result2: 0
real 0m0.452s
user 0m0.049s
sys 0m0.384s
$ time ./t13 100000
running one kernel
running 100000000 loops
result1: 100000000
result2: 0
real 0m4.236s
user 0m2.620s
sys 0m1.599s
$ time ./t13 100000 1
running two kernels
running 100000000 loops
result1: 100000000
result2: 100000000
real 0m4.199s
user 0m2.673s
sys 0m1.514s
$
(CUDA 11.4, CentOS 7, Tesla V100, driver 470.57.02)
I conclude from the above test case that the two kernels are running concurrently, i.e. overlapped.
Note: Although I happened to have CL_QUEUE_OUT_OF_ORDER_EXEC_MODE_ENABLE
set on both queues in the above example, that isn’t necessary for this simple demonstration, and you’ll get a similar result if you replace those flags with 0.