What is the difference between creating a buffer object with clCreateBuffer + CL_MEM_COPY_HOST_PTR vs. clCreateBuffer + clEnqueueWriteBuffer?

Question

I have seen both versions in tutorials, but I could not find out, what their advantages and disadvantages are. Which one is the proper one?

cl_mem input = clCreateBuffer(context,CL_MEM_READ_ONLY,sizeof(float) * DATA_SIZE, NULL, NULL);
clEnqueueWriteBuffer(command_queue, input, CL_TRUE, 0, sizeof(float) * DATA_SIZE, inputdata, 0, NULL, NULL);

vs.

cl_mem input = clCreateBuffer(context,CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR, ,sizeof(float) * DATA_SIZE, inputdata, NULL);

Thanks.

[Update]

I added CL_MEM_COPY_HOST_PTR, to the second example to make it correct.

Answer 1

During my working with OpenCL I found a very important difference between

cl_mem CT = clCreateImage3DContext, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR , Volume_format, X, Y, Z, rowPitch, slicePitch, sourceData, &error);

and

cl_mem CT = clCreateImage3D(Context, CL_MEM_READ_ONLY , Volume_format, X, Y, Z, 0, 0, 0, &error);
error = clEnqueueWriteImage(CommandQue, CT, CL_TRUE, origin, region, rowPitch, slicePitch, sourceData, 0, 0, 0);

For the first approach OpenCL will copy the host pointer not direct to the GPU. First it will allocate a second temporary buffer on the host which can cause problems if you load big stuff like a CT to the GPU. For a short time the needed memory is twice the CT size. Also the data is not copied during this function. It is copied during the argument setting to the kernel function which uses the 3D image object.

The second approach direct copies the data to the GPU. There are no additional allocations done by OpenCL. I think this is probably the same for normal buffer objects.