C++ Nvidia NPP nppiFilter与2d内核进行卷积时会产生垃圾
提供C++ Nvidia NPP nppiFilter与2d内核进行卷积时会产生垃圾,c++,image-processing,cuda,convolution,npp,C++,Image Processing,Cuda,Convolution,Npp,提供nppiFilter函数,用于将用户提供的映像与用户提供的内核进行卷积。对于一维卷积核,nppiFilter工作正常。但是,nppiFilter正在为2D内核生成垃圾映像 我使用典型的Lena图像作为输入: 下面是我用1D卷积内核进行的实验,它可以产生良好的输出 #include <npp.h> // provided in CUDA SDK #include <ImagesCPU.h> // these image libraries are also in C
nppiFilter
函数,用于将用户提供的映像与用户提供的内核进行卷积。对于一维卷积核,nppiFilter
工作正常。但是,nppiFilter
正在为2D内核生成垃圾映像
我使用典型的Lena图像作为输入:
下面是我用1D卷积内核进行的实验,它可以产生良好的输出
#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);
}
下面是使用2D内核的输出图像[-1 0 1;-1 0 1;-1 0 1]
我做错了什么?
描述了一个类似的问题,如用户Steenstrup的图像所示:
最后几点注意事项:
- 对于2D内核,对于某些锚定值(例如,
或nppipointoanchor={0,0}
),我得到error{1,1}
,根据。这一问题在报告中简要提到李>-24
- 这段代码非常冗长。这不是主要问题,但是有人对如何使代码更简洁有什么建议吗
//1D instead of 2D
cudaMalloc((void**)&deviceKernel, kernelSize.width * kernelSize.height * sizeof(Npp32s));
cudaMemcpy(deviceKernel, hostKernel, kernelSize.width * kernelSize.height * sizeof(Npp32s), cudaMemcpyHostToDevice);
另一方面,“比例因子”为1并不表示没有缩放。比例因子为2^(-ScaleFactor)。啊,太好了。我现在正在尝试1D
cudamaloc
和1DcudaMemcpy
。另外,听起来好像ScaleFactor=0
不会给出缩放,对吗?使用1D malloc和memcpy解决了这个问题!!谢谢这是使用2d 3x3内核处理的图像:如果NPP按2^(-ScaleFactor)
缩放,那么我认为ScaleFactor=0
应该给出1的除数。但是,设置ScaleFactor=0
会给我一个空白图像。
//1D instead of 2D
cudaMalloc((void**)&deviceKernel, kernelSize.width * kernelSize.height * sizeof(Npp32s));
cudaMemcpy(deviceKernel, hostKernel, kernelSize.width * kernelSize.height * sizeof(Npp32s), cudaMemcpyHostToDevice);