当尺寸不是2的幂时减去向量的和？ [英] reduction for sum of vector when size is not power of 2?

问题描述

对于GPU上的经典缩减算法，如果向量的大小是2的幂，那么它可以很好地工作。如果不是这样，怎么办？在某一点上，我们将必须找到元素的奇数的和。

解决方案

您可以计算一个没有大小的矩阵的和的权力为二。查看示例：

  #include< math.h> 
 #define N 1022 //总大小
 __global__ void sum（int * A，int * C）
 {
 __shared__ int temp [blockDim.x]; 
 int idx = threadIdx.x + blockDim.x * blockIdx.x; 
 int local_idx = threadIdx.x; 
 temp [local_idx] = A [idx]; 
 int i = ceil（blockDim.x / 2）; 
 __syncthreads（）; 
 while（i！= 0）
 {
 if（idx + i  temp [local_idx] + = tmp [local_idx + i ]; 
 i / = 2; 
 __syncthreads（）; 
 
} 
 if（local_idx == 0）
 C [blockIdx.x] = temp [0]; 
} 
  

For the classical reduction algorithm on GPU, it works perfectly if the size of vector is the power of 2. What if it is not the case? At some point we will have to find the sum of odd number of element. What is the best way to deal with that?

解决方案

You can compute the sum of a matrix that doesn't have a size of a power of two. Look at the example :

#include <math.h>
#define N 1022 //total size
__global__ void sum(int *A, int *C)
{
        __shared__ int temp[blockDim.x];
        int idx = threadIdx.x+blockDim.x*blockIdx.x;
        int local_idx = threadIdx.x;
        temp[local_idx] = A[idx];
        int i=ceil(blockDim.x/2);
        __syncthreads();
        while(i!=0)
        {
                 if(idx+i<N && local_idx<i)
                          temp[local_idx] += tmp[local_idx+i];
                 i/=2;
                 __syncthreads();

        }
       if(local_idx == 0)
           C[blockIdx.x] = temp[0]; 
}

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