如何通过OpenCL C++绑定获取最大全局工作尺寸？

Got it, you’re looking to take full control over the global work size for your OpenCL kernel instead of letting the runtime auto-select it (which might not pick the absolute maximum possible value). Here’s a straightforward way to make that happen:

• First, query the maximum allowed global work size
  The CL_KERNEL_GLOBAL_WORK_SIZE parameter in clGetKernelWorkGroupInfo is exactly what you need here—it lets you fetch the largest global work size supported for your specific kernel and target device. This value is kernel/device-specific, so it’s important to query it for your setup.

  Example code for the query:

  size_t max_global_size[3]; // OpenCL supports up to 3 dimensions
  cl_int err = clGetKernelWorkGroupInfo(kernel, device, CL_KERNEL_GLOBAL_WORK_SIZE, sizeof(max_global_size), max_global_size, NULL);
  if (err != CL_SUCCESS) {
      // Don't forget to handle errors here!
  }
  

• Then, use the queried size in your kernel enqueue call
  Once you have the maximum global size, pass it directly to clEnqueueNDRangeKernel instead of relying on auto-selection. For a 1D kernel, your code would look like this (matching the example you provided, but using the queried max value):

  size_t global_size = max_global_size[0]; // Grab the 1D maximum
  cl_int err = clEnqueueNDRangeKernel(queue, kernel, 1, NULL, &global_size, NULL, 0, NULL, NULL);
  if (err != CL_SUCCESS) {
      // Handle any enqueue errors here
  }
  

  The NULL for the local work size means OpenCL will choose a compatible local size for you, but you can also specify a custom local size if you have specific performance or alignment needs—just make sure it’s compatible with both the global size and device constraints.

• A quick heads-up
  While using the absolute maximum global work size is possible, it’s not always the most performant choice. You might want to benchmark with slightly smaller sizes to find the sweet spot for your workload. But if your goal is strictly to use the largest allowed size, this method works perfectly.

内容的提问来源于stack exchange，提问作者LeMoussel