/* * Copyright 1993-2015 NVIDIA Corporation. All rights reserved. * * Please refer to the NVIDIA end user license agreement (EULA) associated * with this source code for terms and conditions that govern your use of * this software. Any use, reproduction, disclosure, or distribution of * this software and related documentation outside the terms of the EULA * is strictly prohibited. * */ // 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 // CUDA runtime #include // helper functions and utilities to work with CUDA #include #include // This kernel computes a standard parallel reduction and evaluates the // time it takes to do that for each block. The timing results are stored // in device memory. __global__ static void timedReduction(const float *input, float *output, clock_t *timer) { // __shared__ float shared[2 * blockDim.x]; extern __shared__ float shared[]; const int tid = threadIdx.x; const int bid = blockIdx.x; if (tid == 0) timer[bid] = clock(); // Copy input. shared[tid] = input[tid]; shared[tid + blockDim.x] = input[tid + blockDim.x]; // Perform reduction to find minimum. for (int d = blockDim.x; d > 0; d /= 2) { __syncthreads(); if (tid < d) { float f0 = shared[tid]; float f1 = shared[tid + d]; if (f1 < f0) { shared[tid] = f1; } } } // Write result. if (tid == 0) output[bid] = shared[0]; __syncthreads(); if (tid == 0) timer[bid+gridDim.x] = clock(); } #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"); // This will pick the best possible CUDA capable device int dev = findCudaDevice(argc, (const char **)argv); float *dinput = NULL; float *doutput = NULL; clock_t *dtimer = NULL; clock_t timer[NUM_BLOCKS * 2]; float input[NUM_THREADS * 2]; for (int i = 0; i < NUM_THREADS * 2; i++) { input[i] = (float)i; } checkCudaErrors(cudaMalloc((void **)&dinput, sizeof(float) * NUM_THREADS * 2)); checkCudaErrors(cudaMalloc((void **)&doutput, sizeof(float) * NUM_BLOCKS)); checkCudaErrors(cudaMalloc((void **)&dtimer, sizeof(clock_t) * NUM_BLOCKS * 2)); checkCudaErrors(cudaMemcpy(dinput, input, sizeof(float) * NUM_THREADS * 2, cudaMemcpyHostToDevice)); timedReduction<<>>(dinput, doutput, dtimer); checkCudaErrors(cudaMemcpy(timer, dtimer, sizeof(clock_t) * NUM_BLOCKS * 2, cudaMemcpyDeviceToHost)); checkCudaErrors(cudaFree(dinput)); checkCudaErrors(cudaFree(doutput)); checkCudaErrors(cudaFree(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; }