我参考How to find the sum of array in CUDA by reduction在代码中使用归约逻辑。
但是它给出了一些错误。我没有弄错，可以请你帮我吗？
所需规格：1.Cuda工具包v6.52.显卡：GTX 210（计算能力1.2）3. Visual Studio 2013
#include<stdio.h>
#include<cuda.h>
#include<malloc.h>
#include<conio.h>
#include<time.h>
#include<windows.h>
#define SIZE 10
#define N 100
__global__ void vectoreAdd(int *d_a, int *d_b, int *d_c)
{
__shared__ int sdata[256];
int i = threadIdx.x + (blockIdx.x*blockDim.x);
sdata[threadIdx.x] = d_a[i];
__syncthreads();
if (i<SIZE)
for (i = 2; i<SIZE; i++)
{
int counter = 0;
for (int j = 2; j<d_a[i]; j++)
{
if (d_a[i] % j == 0)
{
counter = 1; break;
}
}
if (counter == 0)
{
d_b[i] = d_a[i];
}
}
// do reduction in shared mem
for (int s = 1; s < blockDim.x; s *= 2)
{
int index = 2 * s * threadIdx.x;;
if (index < blockDim.x)
{
sdata[index] += sdata[index + s];
}
__syncthreads();
}
// write result for this block to global mem
if (threadIdx.x == 0)
atomicAdd(d_c, sdata[0]);
}
}
int main()
{
clock_t tic = clock();
int *a, *b, *summation=0, sum = 0,count=-1; //declare summation as double/long if needed
int *d_a, *d_b, *d_c;
//int blocks, block_size = 512;
int size = N * sizeof(int);
a = (int *)malloc(SIZE*sizeof(int));
b = (int *)malloc(SIZE*sizeof(int));
summation = (int *)malloc(SIZE*sizeof(int));
cudaMalloc((void**)&d_a, SIZE * sizeof(int));
cudaMalloc((void**)&d_b, SIZE * sizeof(int));
cudaMalloc((void**)&d_c, SIZE * sizeof(int));
for (int i = 1; i<SIZE; i++)
{
a[i] = i;
b[i] = 0;
}
cudaMemcpy(d_a, a, SIZE*sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(d_b, b, SIZE*sizeof(int), cudaMemcpyHostToDevice);
cudaMemcpy(d_c, c, SIZE*sizeof(int), cudaMemcpyHostToDevice);
/*blocks = SIZE / block_size;
if (SIZE% block_size != 0)
blocks++; */
dim3 blocksize(256); // create 1D threadblock
dim3 gridsize(N / blocksize.x); //create 1D grid
vectoreAdd << < gridsize, blocksize >> >(d_a, d_b, d_c);
//cudaThreadSynchronize();
cudaMemcpy(b, d_b, SIZE*sizeof(int), cudaMemcpyDeviceToHost);
cudaMemcpy(summation, d_c, SIZE*sizeof(int), cudaMemcpyDeviceToHost);
for (int m = 0; m < SIZE; m++)
{
if (b[m] != 0)
{
printf("\n prime no is:%d", b[m]);
count = count + 1;
}
}
printf("\n\n Total prime no. are: %d", count);
/* for (int j = 1; j<SIZE; j++)
{
sum = sum + b[j];
}*/
printf("\n \nsum of all prime no upto %d is:%d", SIZE, summation);
clock_t toc = clock();
printf("\n\nElapsed: %f seconds\n", (double)(toc - tic) / CLOCKS_PER_SEC);
free(a); free(b); free(summation);
cudaFree(d_a); cudaFree(d_b); cudaFree(d_c);
getchar(); return 0;
}
您的代码中有很多错误：
cudaMalloc（（（void **）＆d_a，SIZE * sizeof（int））;
应该是：
cudaMalloc（（void **）＆d_a，N * sizeof（int））; //或
cudaMalloc（（void **）＆d_a，size）;
您已经计算过但没有通过。如果是malloc（），则相同//主机代码