相当于在CUDA中找到连续索引的OpenCl

Question

In CUDA to cover multiple blocks, and thus incerase the range of indices for arrays we do some thing like this:

Host side Code:

 dim3 dimgrid(9,1)// total 9 blocks will be launched    
 dim3 dimBlock(16,1)// each block is having 16 threads  // total no. of threads in  
                   //   the grid is thus 16 x9= 144.        

Device side code

 ...
 ...     
 idx=blockIdx.x*blockDim.x+threadIdx.x;// idx will range from 0 to 143 
 a[idx]=a[idx]*a[idx];
 ...
 ...    

What is the equivalent in OpenCL for acheiving the above case ?

Answer 1

On the host, when you enqueue your kernel using clEnqueueNDRangeKernel , you have to specify the global and local work size. For instance:

size_t global_work_size[1] = { 144 }; // 16 * 9 == 144
size_t local_work_size[1] = { 16 };
clEnqueueNDRangeKernel(cmd_queue, kernel, 1, NULL,
                       global_work_size, local_work_size,
                       0, NULL, NULL);

In your kernel, use:

size_t get_global_size(uint dim);
size_t get_global_id(uint dim);
size_t get_local_size(uint dim);
size_t get_local_id(uint dim);

to retrieve the global and local work sizes and indices respectively, where dim is 0 for x , 1 for y and 2 for z .

The equivalent of your idx will thus be simply size_t idx = get_global_id(0);

See the OpenCL Reference Pages .

Answer 2

Equivalences between CUDA and OpenCL are:

blockIdx.x*blockDim.x+threadIdx.x = get_global_id(0)

LocalSize = blockDim.x

GlobalSize = blockDim.x * gridDim.x