[英]Sum reduction with CUDA: What is N?
根據NVIDIA, 這是最快的減少內核:
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];
}
但是,我不明白“n”參數。 有線索嗎? 我不認為減少數組的大小,因為在while循環中會有緩沖區溢出。
我相信你在幻燈片中發現了一個拼寫錯誤(它應該像while(i + blockDim.x < n)
)。
如果你看一下CUDA SDK示例“ reduce6
”中的源代碼,最新的reduce6
看起來像這樣:
template <class T, unsigned int blockSize, bool nIsPow2>
__global__ void
reduce6(T *g_idata, T *g_odata, unsigned int n)
{
T *sdata = SharedMemory<T>();
// perform first level of reduction,
// reading from global memory, writing to shared memory
...
T mySum = 0;
// we reduce multiple elements per thread. The number is determined by the
// number of active thread blocks (via gridDim). More blocks will result
// in a larger gridSize and therefore fewer elements per thread
while (i < n)
{
mySum += g_idata[i];
// ensure we don't read out of bounds -- this is optimized away for powerOf2 sized arrays
if (nIsPow2 || i + blockSize < n)
mySum += g_idata[i+blockSize];
i += gridSize;
}
請注意while
內的顯式檢查while
以防止越界訪問g_idata
。 你最初的懷疑是正確的; n
只是g_idata
數組的大小。
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