为什么我在 CUDA 中实施总和减少时得到错误的结果?

你真的应该为这样的问题提供一个完整的代码。 您还应该使用正确的 CUDA 错误检查并使用cuda-memcheck运行代码。代码中至少有 2 个错误：

1. 我们不会做这样的cudaMemcpy：

   cudaMemcpy((void**)&d_data, h_data, num_elements * sizeof(float), cudaMemcpyHostToDevice );
   

   它应该是：

   cudaMemcpy(d_data, h_data, num_elements * sizeof(float), cudaMemcpyHostToDevice );
   

   第一个参数只是一个指针，而不是指针到指针。cuda-memcheck或适当的 CUDA 错误检查会将您的注意力集中在这一行上。

2. 您没有启动足够的线程。 内核为每个线程加载一个元素。 如果您的问题大小为 512，您将需要 512 个线程，如下所示：

   int threads = (num_elements/2) + num_elements%2;
   

   不是让你明白的。 不知道你在那里有什么想法。 但这可以为 512 案例修复它：

   int threads = (num_elements);
   

   您的缩减方法需要 2 次幂线程块大小。

这是一个完整的测试用例，请注意cuda-memcheck的使用：

$ cat t27.cu
#include <stdio.h>
#define NUM_ELEMENTS 512
__global__ void reduction(float *g_data, int n)
{
__shared__ float s_data[NUM_ELEMENTS];
int tid = threadIdx.x;
int index = tid + blockIdx.x*blockDim.x;
s_data[tid] = 0.0;
if (index < n){
s_data[tid] = g_data[index];
}
__syncthreads();
for (int s = 2; s <= blockDim.x; s = s * 2){
if ((tid%s) == 0){
s_data[tid] += s_data[tid + s / 2];
}
__syncthreads();
}
if (tid == 0){
g_data[blockIdx.x] = s_data[tid];
}
}
float computeOnDevice(float* h_data, int num_elements)
{
float* d_data = NULL;
float result;
// Memory allocation on device side
cudaMalloc((void**)&d_data, sizeof(float)*num_elements);
// Copy from host memory to device memory
cudaMemcpy(d_data, h_data, num_elements * sizeof(float), cudaMemcpyHostToDevice );
int threads = (num_elements);
// Invoke the kernel
reduction<<< 1 ,threads >>>(d_data,num_elements);
// Copy from device memory back to host memory
cudaMemcpy(&result, d_data, sizeof(float), cudaMemcpyDeviceToHost);
cudaFree(d_data);
cudaDeviceReset();
return result;
}

int main(){
float *data = new float[NUM_ELEMENTS];
for (int i = 0; i < NUM_ELEMENTS; i++) data[i] = 1;
float r = computeOnDevice(data, NUM_ELEMENTS);
printf(" result = %fn" , r);
}
$ nvcc -arch=sm_35 -o t27 t27.cu
$ cuda-memcheck ./t27
========= CUDA-MEMCHECK
result = 512.000000
========= ERROR SUMMARY: 0 errors

这是您现在发布的代码的修改版本(以几种新的/不同的方式损坏)，它似乎对我来说可以正常运行：

$ cat t30.cu
#define NUM_ELEMENTS 512
__global__ void reduction(float *g_data, int n)
{
__shared__ float s_data[NUM_ELEMENTS];
int tid = threadIdx.x;
int index = tid + blockIdx.x*blockDim.x;
s_data[tid] = 0.0;
if (index < n){
s_data[tid] = g_data[index];
}
__syncthreads();
for (int s = 2; s <= blockDim.x; s = s * 2){
if ((tid%s) == 0){
s_data[tid] += s_data[tid + s / 2];
}
__syncthreads();
}
if (tid == 0){
g_data[blockIdx.x] = s_data[tid];
}
}

// includes, system
#include <cuda_runtime.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <float.h>
// includes, kernels
// For simplicity, just to get the idea in this MP, we're fixing the problem size to 512 elements.
#define NUM_ELEMENTS 512
////////////////////////////////////////////////////////////////////////////////
// declaration, forward
void runTest( int argc, char** argv);
float computeOnDevice(float* h_data, int array_mem_size);
extern "C"
void computeGold( float* reference, float* idata, const unsigned int len)
{
for (int i = 0; i<len; i++) *reference += idata[i];
};
////////////////////////////////////////////////////////////////////////////////
// Program main
////////////////////////////////////////////////////////////////////////////////
int main( int argc, char** argv)
{
cudaSetDevice(0);
runTest( argc, argv);
return EXIT_SUCCESS;
}
////////////////////////////////////////////////////////////////////////////////
//! Run naive scan test
////////////////////////////////////////////////////////////////////////////////
void runTest( int argc, char** argv)
{
int num_elements = NUM_ELEMENTS;
const unsigned int array_mem_size = sizeof( float) * num_elements;
// Allocate host memory to store the input data
float* h_data = (float*) malloc( array_mem_size);
// initialize the input data on the host to be integer values
// between 0 and 1000
for( unsigned int i = 0; i < num_elements; ++i)
h_data[i] = floorf(1000*(rand()/(float)RAND_MAX));
// Function to compute the reference solution on CPU using a C sequential version of the algorithm
// It is written in the file "vector_reduction_gold.cpp". The Makefile compiles this file too.
float reference = 0.0f;
computeGold(&reference , h_data, num_elements);
// Function to compute the solution on GPU using a call to a CUDA kernel (see body below)
// The kernel is written in the file "vector_reduction_kernel.cu". The Makefile also compiles this file.
float result = computeOnDevice(h_data, num_elements);
// We can use an epsilon of 0 since values are integral and in a range that can be exactly represented
float epsilon = 0.0f;
unsigned int result_regtest = (abs(result - reference) <= epsilon);
printf( "Test %sn", (1 == result_regtest) ? "CORRECTO: Coinciden los resultados de la CPU y la GPU" : "INCORRECTO: Los resultados calculados en paralelo en la GPU no coinciden con los obtenidos secuencialmente en la CPU");
printf( "device: %f  host: %fn", result, reference);
// cleanup memory
free( h_data);
}
// Function to call the CUDA kernel on the GPU.
// Take h_data from host, copies it to device, setup grid and thread
// dimensions, excutes kernel function, and copy result of scan back
// to h_data.
// Note: float* h_data is both the input and the output of this function.
#if 0
float computeOnDevice(float* h_data, int num_elements)
{
float* d_data = NULL;
float result;
// Memory allocation on device side
cudaMalloc((void**)&d_data, sizeof(float)*num_elements);
// Copy from host memory to device memory
cudaMemcpy((void**)&d_data, h_data, num_elements * sizeof(float), cudaMemcpyHostToDevice );
//int threads = (num_elements/2) + num_elements%2;
int threads = (num_elements);
// Invoke the kernel
reduction<<< 1 ,threads >>>(d_data,num_elements);
// Copy from device memory back to host memory
cudaMemcpy(&result, d_data, sizeof(float), cudaMemcpyDeviceToHost);
cudaFree(d_data);
cudaDeviceReset();
return result;
}
#endif
float computeOnDevice(float* h_data, int num_elements)
{
float* d_data = NULL;
float result;
// Memory allocation on device side
cudaError_t err = cudaMalloc((void**)&d_data, sizeof(float)*num_elements);
if (err != cudaSuccess) {printf("CUDA error: %sn", cudaGetErrorString(err)); exit(0);}
// Copy from host memory to device memory
cudaMemcpy(d_data, h_data, num_elements * sizeof(float), cudaMemcpyHostToDevice );
int threads = (num_elements);
// Invoke the kernel
reduction<<< 1 ,threads >>>(d_data,num_elements);
// Copy from device memory back to host memory
cudaMemcpy(&result, d_data, sizeof(float), cudaMemcpyDeviceToHost);
cudaFree(d_data);
err = cudaGetLastError();
if (err != cudaSuccess) {printf("CUDA error: %sn", cudaGetErrorString(err)); exit(0);}
cudaDeviceReset();
return result;
}
$ nvcc -arch=sm_35 -o t30 t30.cu
$ cuda-memcheck ./t30
========= CUDA-MEMCHECK
Test CORRECTO: Coinciden los resultados de la CPU y la GPU
device: 260795.000000  host: 260795.000000
========= ERROR SUMMARY: 0 errors
$

您仍然没有将适当的 CUDA 错误检查添加到您的代码中，因此您完全有可能遇到机器设置问题。 如果您仍然遇到问题，您可能需要运行我上面发布的确切代码，因为我在其中进行了基本的错误检查。