According to NVIDIA, this is the fastest sum reduction kernel:
template <unsigned int blockSize>
__device__ void warpReduce(volatile int *sdata, unsigned int tid) {
if (blockSize >= 64) sdata[tid] += sdata[tid + 32];
if (blockSize >= 32) sdata[tid] += sdata[tid + 16];
if (blockSize >= 16) sdata[tid] += sdata[tid + 8];
if (blockSize >= 8) sdata[tid] += sdata[tid + 4];
if (blockSize >= 4) sdata[tid] += sdata[tid + 2];
if (blockSize >= 2) sdata[tid] += sdata[tid + 1];
}
template <unsigned int blockSize>
__global__ void reduce6(int *g_idata, int *g_odata, unsigned int n) {
extern __shared__ int sdata[];
unsigned int tid = threadIdx.x;
unsigned int i = blockIdx.x*(blockSize*2) + tid;
unsigned int gridSize = blockSize*2*gridDim.x;
sdata[tid] = 0;
while (i < n) { sdata[tid] += g_idata[i] + g_idata[i+blockSize]; i += gridSize; }
__syncthreads();
if (blockSize >= 512) { if (tid < 256) { sdata[tid] += sdata[tid + 256]; } __syncthreads(); }
if (blockSize >= 256) { if (tid < 128) { sdata[tid] += sdata[tid + 128]; } __syncthreads(); }
if (blockSize >= 128) { if (tid < 64) { sdata[tid] += sdata[tid + 64]; } __syncthreads(); }
if (tid < 32) warpReduce(sdata, tid);
if (tid == 0) g_odata[blockIdx.x] = sdata[0];
}
However, I don't understand the "n" parameter. Any clues? I don't think it's the size of the array to reduce, since in the while loop there would be a buffer overflow.