Nvidia 性能基元 (NPP) http://developer.nvidia.com/cuda/nvidia-performance-primitives提供了nppiFilter用于将用户提供的图像与用户提供的内核进行卷积的函数。对于一维卷积核，nppiFilter工作正常。然而，nppiFilter正在为 2D 内核生成垃圾图像。

I used the typical Lena image as input:

这是我的实验1D卷积核，产生良好的输出。

#include <npp.h> // provided in CUDA SDK
#include <ImagesCPU.h> // these image libraries are also in CUDA SDK
#include <ImagesNPP.h>
#include <ImageIO.h>

void test_nppiFilter()
{
    npp::ImageCPU_8u_C1 oHostSrc;
    npp::loadImage("Lena.pgm", oHostSrc);
    npp::ImageNPP_8u_C1 oDeviceSrc(oHostSrc); // malloc and memcpy to GPU 
    NppiSize kernelSize = {3, 1}; // dimensions of convolution kernel (filter)
    NppiSize oSizeROI = {oHostSrc.width() - kernelSize.width + 1, oHostSrc.height() - kernelSize.height + 1};
    npp::ImageNPP_8u_C1 oDeviceDst(oSizeROI.width, oSizeROI.height); // allocate device image of appropriately reduced size
    npp::ImageCPU_8u_C1 oHostDst(oDeviceDst.size());
    NppiPoint oAnchor = {2, 1}; // found that oAnchor = {2,1} or {3,1} works for kernel [-1 0 1] 
    NppStatus eStatusNPP;

    Npp32s hostKernel[3] = {-1, 0, 1}; // convolving with this should do edge detection
    Npp32s* deviceKernel;
    size_t deviceKernelPitch;
    cudaMallocPitch((void**)&deviceKernel, &deviceKernelPitch, kernelSize.width*sizeof(Npp32s), kernelSize.height*sizeof(Npp32s));
    cudaMemcpy2D(deviceKernel, deviceKernelPitch, hostKernel,
                     sizeof(Npp32s)*kernelSize.width, // sPitch
                     sizeof(Npp32s)*kernelSize.width, // width
                     kernelSize.height, // height
                     cudaMemcpyHostToDevice);
    Npp32s divisor = 1; // no scaling

    eStatusNPP = nppiFilter_8u_C1R(oDeviceSrc.data(), oDeviceSrc.pitch(),
                                          oDeviceDst.data(), oDeviceDst.pitch(),
                                          oSizeROI, deviceKernel, kernelSize, oAnchor, divisor);

    cout << "NppiFilter error status " << eStatusNPP << endl; // prints 0 (no errors)
    oDeviceDst.copyTo(oHostDst.data(), oHostDst.pitch()); // memcpy to host
    saveImage("Lena_filter_1d.pgm", oHostDst); 
}

Output of the above code with kernel [-1 0 1] -- it looks like a reasonable gradient image:

然而，nppiFilter如果我使用，输出垃圾图像2D卷积核。以下是我为使用 2D 内核运行而对上述代码进行的更改[-1 0 1; -1 0 1; -1 0 1]:

NppiSize kernelSize = {3, 3};
Npp32s hostKernel[9] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
NppiPoint oAnchor = {2, 2}; // note: using anchor {1,1} or {0,0} causes error -24 (NPP_TEXTURE_BIND_ERROR)
saveImage("Lena_filter_2d.pgm", oHostDst);

下面是使用 2D 内核的输出图像[-1 0 1; -1 0 1; -1 0 1].

我究竟做错了什么？

这篇 StackOverflow 帖子 https://stackoverflow.com/questions/12778463/cuda-npp-filters描述了类似的问题，如用户 Steensrup 的图片所示：http://1ordrup.dk/kasper/image/Lena_boxFilter5.jpg http://1ordrup.dk/kasper/image/Lena_boxFilter5.jpg

最后一些注意事项：

• 对于 2D 内核，对于某些锚点值（例如NppiPoint oAnchor = {0, 0} or {1, 1}），我收到错误-24，这意味着NPP_TEXTURE_BIND_ERROR根据核电站用户指南 http://developer.download.nvidia.com/compute/DevZone/docs/html/CUDALibraries/doc/NPP_Library.pdf。这个问题在这个 StackOverflow 帖子 https://stackoverflow.com/questions/12778463/cuda-npp-filters.
• 这段代码非常冗长。这不是主要问题，但是有人对如何使这段代码更加简洁有任何建议吗？

您正在为内核数组使用 2D 内存分配器。内核数组是密集的一维数组，而不是典型 NPP 图像那样的二维跨步数组。

只需将 2D CUDA malloc 替换为大小为 kernelWidth*kernelHeight*sizeof(Npp32s) 的简单 cuda malloc，并执行正常的 CUDA memcopy 而不是 memcopy 2D。

//1D instead of 2D
cudaMalloc((void**)&deviceKernel, kernelSize.width * kernelSize.height * sizeof(Npp32s));
cudaMemcpy(deviceKernel, hostKernel, kernelSize.width * kernelSize.height * sizeof(Npp32s), cudaMemcpyHostToDevice);

顺便说一句，“比例因子”1 并不意味着不缩放。缩放系数为 2^(-ScaleFactor)。