/* Copyright (c) 2022, NVIDIA CORPORATION. All rights reserved. * * Redistribution and use in source and binary forms, with or without * modification, are permitted provided that the following conditions * are met: * * Redistributions of source code must retain the above copyright * notice, this list of conditions and the following disclaimer. * * Redistributions in binary form must reproduce the above copyright * notice, this list of conditions and the following disclaimer in the * documentation and/or other materials provided with the distribution. * * Neither the name of NVIDIA CORPORATION nor the names of its * contributors may be used to endorse or promote products derived * from this software without specific prior written permission. * * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS ``AS IS'' AND ANY * EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR * PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR * CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, * EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, * PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR * PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY * OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT * (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE * OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. */ /* * This example shows how to use the clock function to measure the performance * of block of threads of a kernel accurately. Blocks are executed in parallel * and out of order. Since there's no synchronization mechanism between blocks, * we measure the clock once for each block. The clock samples are written to * device memory. */ // System includes #include #include #include #include #include // helper functions and utilities to work with CUDA #include #define NUM_BLOCKS 64 #define NUM_THREADS 256 // It's interesting to change the number of blocks and the number of threads to // understand how to keep the hardware busy. // // Here are some numbers I get on my G80: // blocks - clocks // 1 - 3096 // 8 - 3232 // 16 - 3364 // 32 - 4615 // 64 - 9981 // // With less than 16 blocks some of the multiprocessors of the device are idle. // With // more than 16 you are using all the multiprocessors, but there's only one // block per // multiprocessor and that doesn't allow you to hide the latency of the memory. // With // more than 32 the speed scales linearly. // Start the main CUDA Sample here int main(int argc, char **argv) { printf("CUDA Clock sample\n"); typedef long clock_t; clock_t timer[NUM_BLOCKS * 2]; float input[NUM_THREADS * 2]; for (int i = 0; i < NUM_THREADS * 2; i++) { input[i] = (float)i; } char *cubin, *kernel_file; size_t cubinSize; kernel_file = sdkFindFilePath("clock_kernel.cu", argv[0]); compileFileToCUBIN(kernel_file, argc, argv, &cubin, &cubinSize, 0); CUmodule module = loadCUBIN(cubin, argc, argv); CUfunction kernel_addr; checkCudaErrors(cuModuleGetFunction(&kernel_addr, module, "timedReduction")); dim3 cudaBlockSize(NUM_THREADS, 1, 1); dim3 cudaGridSize(NUM_BLOCKS, 1, 1); CUdeviceptr dinput, doutput, dtimer; checkCudaErrors(cuMemAlloc(&dinput, sizeof(float) * NUM_THREADS * 2)); checkCudaErrors(cuMemAlloc(&doutput, sizeof(float) * NUM_BLOCKS)); checkCudaErrors(cuMemAlloc(&dtimer, sizeof(clock_t) * NUM_BLOCKS * 2)); checkCudaErrors(cuMemcpyHtoD(dinput, input, sizeof(float) * NUM_THREADS * 2)); void *arr[] = {(void *)&dinput, (void *)&doutput, (void *)&dtimer}; checkCudaErrors(cuLaunchKernel( kernel_addr, cudaGridSize.x, cudaGridSize.y, cudaGridSize.z, /* grid dim */ cudaBlockSize.x, cudaBlockSize.y, cudaBlockSize.z, /* block dim */ sizeof(float) * 2 * NUM_THREADS, 0, /* shared mem, stream */ &arr[0], /* arguments */ 0)); checkCudaErrors(cuCtxSynchronize()); checkCudaErrors( cuMemcpyDtoH(timer, dtimer, sizeof(clock_t) * NUM_BLOCKS * 2)); checkCudaErrors(cuMemFree(dinput)); checkCudaErrors(cuMemFree(doutput)); checkCudaErrors(cuMemFree(dtimer)); long double avgElapsedClocks = 0; for (int i = 0; i < NUM_BLOCKS; i++) { avgElapsedClocks += (long double)(timer[i + NUM_BLOCKS] - timer[i]); } avgElapsedClocks = avgElapsedClocks / NUM_BLOCKS; printf("Average clocks/block = %Lf\n", avgElapsedClocks); return EXIT_SUCCESS; }