我使用此代码进行简化：
http://www.math.nsysu.edu.tw/~lam/MPI/code/cuda/reduction.cu
基于Mark Harris的谈话，如此处
http://www.math.nsysu.edu.tw/~lam/MPI/lecture/reduction.pdf
但是
TryGetValue内核reduce6起作用，而reduce7失败。 bcos reduce7是否取决于大小必须达到上面定义的“大小”的共享内存量？
代码段在这里：
#define blocksize 1024
#define gridsize 1024*8
#define size blocksize*gridsize
这个内核从main像这样被调用：
#define THR_PER_BLC 1024
#define BLC_PER_GRD 16
#define GRID_SIZE THR_PER_BLC * BLC_PER_GRD
template<unsigned int nThreads>
__global__ void reduce7(int *g_idata, int *g_odata, unsigned int n) {
//I added GRID_SIZE myself so it can be volatile
__shared__ volatile int sdata[THR_PER_BLC];
unsigned int tid = threadIdx.x;
unsigned int i = blockIdx.x * (nThreads * 2) + threadIdx.x;
unsigned int gridSize = nThreads * 2 * gridDim.x;
sdata[tid] = 0;
while (i < n) {
sdata[tid] += g_idata[i] + g_idata[i + nThreads];
i += gridSize;
}
__syncthreads();
// reduction in shared memory
if (nThreads >= 512) {
if (tid < 256) { sdata[tid] += sdata[tid + 256]; }
__syncthreads();
}
if (nThreads >= 256) {
if (tid < 128) { sdata[tid] += sdata[tid + 128]; }
__syncthreads();
}
if (nThreads >= 128) {
if (tid < 64) { sdata[tid] += sdata[tid + 64]; }
__syncthreads();
}
if (tid < 32) {
if (nThreads >= 64) sdata[tid] += sdata[tid + 32];
if (nThreads >= 32) sdata[tid] += sdata[tid + 16];
if (nThreads >= 16) sdata[tid] += sdata[tid + 8];
if (nThreads >= 8) sdata[tid] += sdata[tid + 4];
if (nThreads >= 4) sdata[tid] += sdata[tid + 2];
if (nThreads >= 2) sdata[tid] += sdata[tid + 1];
// transfer of the result to global memory
if (tid == 0) g_odata[blockIdx.x] = sdata[0];
}
}
从根本上讲，不能对大GRID_SIZE调用reduce7吗？
这是我的测试
threads = THR_PER_BLC /2 ;
int gsize = BLC_PER_GRD /8;
switch (threads) {
case 512:
reduce7<512> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 256:
reduce7<256> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 128:
reduce7<128> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 64:
reduce7<64> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 32:
reduce7<32> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 16:
reduce7<16> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 8:
reduce7<8> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 4:
reduce7<4> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 2:
reduce7<2> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
case 1:
reduce7<1> << < gsize, threads >> > (g_idata, g_odata, GRID_SIZE);
break;
}
cudaThreadSynchronize();
还有我的gpu：
#################################################################
6 Unroll the complete loop
Kernal elapsed time = 0.030(ms)
Elapsed time = 0.057(ms)
Sum = 8192, with BLC_PER_GRD 16 THR_PER_BLC 512
#################################################################
7 Final
Kernal elapsed time = 0.015(ms), band =
Elapsed time = 0.040(ms)
Sum = 8192, with BLC_PER_GRD 16 THR_PER_BLC 512
#################################################################
#################################################################
6 Unroll the complete loop
Kernal elapsed time = 0.031(ms)
Elapsed time = 0.057(ms)
Sum = 8192, with BLC_PER_GRD 8 THR_PER_BLC 1024
#################################################################
7 Final
Kernal elapsed time = 0.015(ms), band =
Elapsed time = 0.040(ms)
Sum = 8192, with BLC_PER_GRD 8 THR_PER_BLC 1024
#################################################################
#################################################################
6 Unroll the complete loop
Kernal elapsed time = 0.569(ms)
Elapsed time = 12.889(ms)
Sum = 8388608, with BLC_PER_GRD 8192 THR_PER_BLC 1024
#################################################################
嗯，让我们先设置128个线程，网格大小为4：
a@M:/usr/local/cuda/samples/bin/x86_64/linux/release$ ./dev*Drv
./deviceQueryDrv Starting...
CUDA Device Query (Driver API) statically linked version
Detected 1 CUDA Capable device(s)
Device 0: "GeForce GTX 1060 6GB"
CUDA Driver Version: 9.2
CUDA Capability Major/Minor version number: 6.1
Total amount of global memory: 6078 MBytes (6373572608 bytes)
(10) Multiprocessors, (128) CUDA Cores/MP: 1280 CUDA Cores
GPU Max Clock rate: 1709 MHz (1.71 GHz)
Memory Clock rate: 4004 Mhz
Memory Bus Width: 192-bit
L2 Cache Size: 1572864 bytes
Max Texture Dimension Sizes 1D=(131072) 2D=(131072, 65536) 3D=(16384, 16384, 16384)
Maximum Layered 1D Texture Size, (num) layers 1D=(32768), 2048 layers
Maximum Layered 2D Texture Size, (num) layers 2D=(32768, 32768), 2048 layers
Total amount of constant memory: 65536 bytes
Total amount of shared memory per block: 49152 bytes
Total number of registers available per block: 65536
Warp size: 32
Maximum number of threads per multiprocessor: 2048
Maximum number of threads per block: 1024
Max dimension size of a thread block (x,y,z): (1024, 1024, 64)
Max dimension size of a grid size (x,y,z): (2147483647, 65535, 65535)
Texture alignment: 512 bytes
Maximum memory pitch: 2147483647 bytes
Concurrent copy and kernel execution: Yes with 2 copy engine(s)
Run time limit on kernels: No
Integrated GPU sharing Host Memory: No
Support host page-locked memory mapping: Yes
Concurrent kernel execution: Yes
Alignment requirement for Surfaces: Yes
Device has ECC support: Disabled
Device supports Unified Addressing (UVA): Yes
Device supports Compute Preemption: Yes
Supports Cooperative Kernel Launch: Yes
Supports MultiDevice Co-op Kernel Launch: Yes
Device PCI Domain ID / Bus ID / location ID: 0 / 3 / 0
Compute Mode:
< Default (multiple host threads can use ::cudaSetDevice() with device simultaneously) >
然后reduce7起作用。因此，意味着reduce7严格取决于max shm？
似乎我对以下行感到困惑：#define MAX_SHM 49152
#define GRID_SIZE MAX_SHM / sizeof(int)
#define THR_PER_BLC 128
#define BLC_PER_GRD GRID_SIZE/THR_PER_BLC
，其中n为GRID_SIZE。那么现在我不知道while (i < n) {是什么意思。需要消化一段时间。但很高兴知道，在一个块中只能有特定数量的线程，在这种情况下，我们必须与SM匹配。
答案 0 :(得分：4)
首先，这种减少所需的共享内存仅与 block 的需求一样大，而不是 grid 的需求。因此，要求按网格大小分配共享内存是没有道理的。
第二，这要求每个块64KB静态分配的共享内存：
__shared__ volatile int sdata[GRID_SIZE];
那是行不通的，因为：
Total amount of shared memory per block: 49152 bytes
此外，此外，这是每个块要求64KB动态分配的共享内存：
case 128:
reduce7<128> << < gsize, threads, GRID_SIZE * sizeof(int) >> > (g_idata, g_odata, GRID_SIZE);
break;
因此组合（64K + 64K）将永远无法工作。
您似乎对如何使用共享内存以及每个块需要多少内存感到困惑。该块每个线程仅需要一个数量（在这种情况下为int。
您可能会对the syntax and usage of statically allocated shared memory vs. dynamically allocated shared memory感到困惑。对于这种类型的问题，通常会使用其中一个，而不是两者都使用。
我不知道这句话的意思：
//I added GRID_SIZE myself so it can be volatile
通常的建议：每当您遇到CUDA代码问题时，您都应该做proper CUDA error checking，并用cuda-memcheck运行代码，之前向他人寻求帮助。即使开始的示例代码没有正确的CUDA错误检查，一旦开始进行修改并遇到麻烦，也应该添加。
然后reduce7起作用。因此，意味着reduce7严格取决于max shm？
这意味着reduce7每个块需要一定数量的共享内存。该数量是每个线程一个int。这就是它所需要的。如果您提供更多，只要您不超过所能提供的最大值，就可以（一种）。如果您超过了可以提供的最大值，则整个内核启动将失败。
换句话说，您真正需要的是：
__shared__ volatile int sdata[THR_PER_BLC];