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好的，所以任务的主要思想是计算多个图像的平均值，我让它以正常方式运行，所以我想我会尝试使用 CUDA，但不幸的是我在输出中收到的是第一张图像而不是平均值。（在内核内部，我还尝试将一些像素设置为 0 以确保正在发生某些事情但没有运气..）

////My kernel:
//nImages - number of images in the memory
//nBytes - number of pixels*color per image (also it's a size of dataOut)
//nImages*nBytes gives us the size of dataIn 
//nBatch - dataIn has 1 milion bytes per image, we run in 6144 threads, so we need 163 batches to calc the whole dataOut
__global__ 
void avg_arrays(unsigned char* cuDataIn, unsigned char* cuDataOut, int nImages, int nBytes, int nBatch) 
{
   //get the position of the correct byte
   int j = threadIdx.x +  nBatch;
   //if we're outside of image then give up
   if(j >= nBytes) return;
   //proceed averaging
   long lSum = 0;
   for(int i=0; i < nImages; ++i) 
      lSum += cuDataIn[i*nBytes + j];
   lSum = lSum / nImages;
   cuDataOut[j] = lSum;
}

内存分配等

unsigned char* dataIn = 0;
unsigned char* dataOut= 0;

// Allocate and Transfer memory to the devicea
gpuErrchk( cudaMalloc((void**)&dataIn, nPixelCountBGR * nNumberOfImages * sizeof(unsigned char)));                                  //dataIn
gpuErrchk( cudaMalloc((void**)&dataOut, nPixelCountBGR * sizeof(unsigned char)));                               //dataOut
gpuErrchk( cudaMemcpy(dataIn, bmps,  nPixelCountBGR * nNumberOfImages * sizeof(unsigned char), cudaMemcpyHostToDevice ));           //dataIn
gpuErrchk( cudaMemcpy(dataOut, basePixels, nPixelCountBGR * sizeof(unsigned char), cudaMemcpyHostToDevice ));   //dataOut

// Perform the array addition
dim3 dimBlock(N);  
dim3 dimGrid(1);

//do it in batches, unless it's possible to run more threads at once, anyway N is a number of max threads
for(int i=0; i<nPixelCountBGR; i+=N){
   cout << "Running with: nImg: "<< nNumberOfImages << ", nPixBGR " << nPixelCountBGR << ", and i = " << i << endl;
   avg_arrays<<<dimGrid, dimBlock>>>(dataIn, dataOut, nNumberOfImages, nPixelCountBGR, 0);
}
// Copy the Contents from the GPU
gpuErrchk(cudaMemcpy(basePixels, dataOut, nPixelCountBGR * sizeof(unsigned char), cudaMemcpyDeviceToHost)); 

gpuErrchk(cudaFree(dataOut));
gpuErrchk(cudaFree(dataIn));

错误检查不会带来任何消息，所有代码运行顺利，最后我得到的只是第一张图像的精确副本。

以防万一有人需要这里的一些控制台输出：

Running with: nImg: 29, nPixBGR 1228800, and i = 0
...
Running with: nImg: 29, nPixBGR 1228800, and i = 1210368
Running with: nImg: 29, nPixBGR 1228800, and i = 1216512
Running with: nImg: 29, nPixBGR 1228800, and i = 1222656
Time of averaging: 0.219
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1 回答 1

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如果N大于 512 或 1024（取决于您在哪个 GPU 上运行，您没有提及），那么这是无效的：

dim3 dimBlock(N);

因为您无法启动每个块超过 512 或 1024 个线程的内核：

 avg_arrays<<<dimGrid, dimBlock>>>(...
                          ^
                          |
                     this is limited to 512 or 1024

如果您研究正确的 cuda 错误检查并将其应用于内核启动，您将捕获此类错误。

于 2013-11-12T16:58:15.710 回答