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我複製了一個異步CUDA/C++示例並對其進行了修改以評估素數。我的問題是，對於每個打印的素數，數組中的下一個值是該值的重複值。這是預期的行爲還是與我編程示例的方式有關？CUDA計算之後在數組中重複值
驗證碼：
////////////////////////////////////////////////////////////////////////////
//
// 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 sample illustrates the usage of CUDA events for both GPU timing and
// overlapping CPU and GPU execution. Events are inserted into a stream
// of CUDA calls. Since CUDA stream calls are asynchronous, the CPU can
// perform computations while GPU is executing (including DMA memcopies
// between the host and device). CPU can query CUDA events to determine
// whether GPU has completed tasks.
//
// includes, system
#include <stdio.h>
// includes CUDA Runtime
#include <cuda_runtime.h>
// includes, project
#include <helper_cuda.h>
#include <helper_functions.h> // helper utility functions
//set matrix to possible prime values
//evaluate if input is prime, sets variable to 0 if not prime
__global__ void testPrimality(int * g_data) {
int idx = blockIdx.x * blockDim.x + threadIdx.x;
g_data[idx] = 3 + idx/2;
if (g_data[idx] <= 3) {
if (g_data[idx] <= 1) {
g_data[idx] = 0;
}
}
else if (g_data[idx] % 2 == 0 || g_data[idx] % 3 == 0) {
g_data[idx] = 0;
}
else {
for (unsigned short i = 5; i * i <= g_data[idx]; i += 6) {
if (g_data[idx] % i == 0 || g_data[idx] % (i + 2) == 0) {
g_data[idx] = 0;
}
}
}
}
bool correct_output(int *data, const int n, const int x)
{
for (int i = 0; i < n; i++)
if (data[i] != x)
{
printf("Error! data[%d] = %d, ref = %d\n", i, data[i], x);
return false;
}
return true;
}
int main(int argc, char *argv[])
{
int devID;
cudaDeviceProp deviceProps;
printf("[%s] - Starting...\n", argv[0]);
// This will pick the best possible CUDA capable device
devID = findCudaDevice(argc, (const char **)argv);
// get device name
checkCudaErrors(cudaGetDeviceProperties(&deviceProps, devID));
printf("CUDA device [%s]\n", deviceProps.name);
const int n = 16 * 1024 * 1024;
int nbytes = n * sizeof(int);
int value = 1;
// allocate host memory
int *a = 0;
checkCudaErrors(cudaMallocHost((void **)&a, nbytes));
memset(a, 0, nbytes);
// allocate device memory
int *d_a=0;
checkCudaErrors(cudaMalloc((void **)&d_a, nbytes));
checkCudaErrors(cudaMemset(d_a, 255, nbytes));
// set kernel launch configuration
dim3 threads = dim3(512, 1);
dim3 blocks = dim3(n/threads.x, 1);
// create cuda event handles
cudaEvent_t start, stop;
checkCudaErrors(cudaEventCreate(&start));
checkCudaErrors(cudaEventCreate(&stop));
StopWatchInterface *timer = NULL;
sdkCreateTimer(&timer);
sdkResetTimer(&timer);
checkCudaErrors(cudaDeviceSynchronize());
float gpu_time = 0.0f;
// asynchronously issue work to the GPU (all to stream 0)
sdkStartTimer(&timer);
cudaEventRecord(start, 0);
cudaMemcpyAsync(d_a, a, nbytes, cudaMemcpyHostToDevice, 0);
//increment_kernel<<<blocks, threads, 0, 0>>>(d_a);
testPrimality<<<blocks, threads, 0, 0 >>>(d_a);
cudaMemcpyAsync(a, d_a, nbytes, cudaMemcpyDeviceToHost, 0);
cudaEventRecord(stop, 0);
sdkStopTimer(&timer);
// have CPU do some work while waiting for stage 1 to finish
unsigned long int counter=0;
while (cudaEventQuery(stop) == cudaErrorNotReady)
{
counter++;
}
checkCudaErrors(cudaEventElapsedTime(&gpu_time, start, stop));
// print the cpu and gpu times
printf("time spent executing by the GPU: %.2f\n", gpu_time);
printf("time spent by CPU in CUDA calls: %.2f\n", sdkGetTimerValue(&timer));
printf("CPU executed %lu iterations while waiting for GPU to finish\n", counter);
//print values for all allocated memory space
for (int i = 0; i < n; i++) {
if (a[i] != 0) {
std::cout << a[i]<< " : " << i << std::endl;
}
}
// check the output for correctness
//bool bFinalResults = correct_output(a, n, value);
bool bFinalResults = true;
// release resources
checkCudaErrors(cudaEventDestroy(start));
checkCudaErrors(cudaEventDestroy(stop));
checkCudaErrors(cudaFreeHost(a));
checkCudaErrors(cudaFree(d_a));
// cudaDeviceReset causes the driver to clean up all state. While
// not mandatory in normal operation, it is good practice. It is also
// needed to ensure correct operation when the application is being
// profiled. Calling cudaDeviceReset causes all profile data to be
// flushed before the application exits
cudaDeviceReset();
exit(bFinalResults ? EXIT_SUCCESS : EXIT_FAILURE);
}
我試圖從整數3開始，對於每個單獨的計算，它應該增加2（因爲素數從來沒有）。我認爲我可以做到這一點，因爲idx似乎總是4的倍數。我想我可以通過將idx除以2來得到2的增量。 – Stephen
然後，也許你想要3+'idx' * 2。 'idx'是你的全局唯一線程索引，它從0開始，每個線程增加1個0,1,2,3，...所以如果你想要一個數值輸入序列3,5,7,9， ...正確的算術是3 +'idx' * 2。 'idx'因爲你有它並不總是4的倍數。 –
我按你的建議做了，但它似乎仍然是重複的價值。 – Stephen