CUDA NPP -打印输出错误

CUDA NPP - Error on printing output

本文关键字：输出 错误 打印 NPP CUDA      更新时间：2023-10-16

遵循我之前的帖子:CUDA NPP - GPU错误检查时出现未知错误

我尝试用CUDA NPP库将图像中的所有像素相加，在一些开发人员的帮助下，我终于编译了我的代码。然而，当我尝试通过将其复制到double变量(与CUDA v4.2的NPP指南一致)来打印存储在partialSum中的值时，我得到了这个错误:

Unhandled exception at 0x00fdf7f4 in MedianFilter.exe: 0xC0000005: Access violation reading location 0x40000000.

我一直在努力摆脱它，但是，到目前为止，我一直没有成功。请帮助!我花了两天的时间研究这一小段代码。

代码: 

#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, char *file, int line, bool abort=true)
{
    if (code != cudaSuccess) 
    {
        fprintf(stderr,"GPUassert: %s %s %dn", cudaGetErrorString(code), file, line);
        if (abort) getchar();
    }
}
// processing image starts here 
// device_pointer initializations
unsigned char *device_input;
unsigned char *device_output;    
size_t d_ipimgSize = input.step * input.rows;
size_t d_opimgSize = output.step * output.rows;
gpuErrchk( cudaMalloc( (void**) &device_input, d_ipimgSize) );
gpuErrchk( cudaMalloc( (void**) &device_output, d_opimgSize) );
gpuErrchk( cudaMemcpy(device_input, input.data, d_ipimgSize, cudaMemcpyHostToDevice) );

// Median filter the input image here
// .......

// allocate data on the host for comparing the sum of all pixels in image with CUDA implementation
// 1st argument - allocate data for pSrc - copy device_output into this pointer
Npp8u *odata; 
gpuErrchk( cudaMalloc( (void**) &odata, sizeof(Npp8u)*output.rows*output.cols ) );
gpuErrchk( cudaMemcpy(odata, device_output, sizeof(Npp8u)*output.rows*output.cols, cudaMemcpyDeviceToDevice) ); 
// 2nd arg - set step 
int ostep = output.step;  
// 3rd arg - set nppiSize
NppiSize imSize; 
imSize.width = output.cols; 
imSize.height = output.rows;
// 4th arg - set npp8u scratch buffer size
Npp8u *scratch; 
int bytes = 0;
nppiReductionGetBufferHostSize_8u_C1R( imSize, &bytes);
gpuErrchk( cudaMalloc((void **)&scratch, bytes) );
// 5th arg - set npp64f partialSum (64 bit double will be the result)
Npp64f *partialSum; 
gpuErrchk( cudaMalloc( (void**) &partialSum, sizeof(Npp64f) ) );
//                 nnp8u, int, nppisize, npp8u, npp64f    
nppiSum_8u_C1R( odata, ostep, imSize, scratch, partialSum );
double *dev_result;
    dev_result = (double*)malloc(sizeof(double)); // EDIT
gpuErrchk( cudaMemcpy(&dev_result, partialSum, sizeof(double), cudaMemcpyDeviceToHost) );
//int tot = output.rows * output.cols;
printf( "n Total Sum cuda %f n",  *dev_result) ;   // <---- access violation here

这里的问题似乎是基本的指针误用(我说似乎是因为我们有不完整的，不可编译的代码，所以很难确定)。

这个应该可以工作:

double *dev_result = (double*)malloc(sizeof(double));
gpuErrchk( cudaMemcpy(dev_result, partialSum, sizeof(double), cudaMemcpyDeviceToHost) );
printf( "n Total Sum cuda %f n",  *dev_result);

这个也可以:

double dev_result;
gpuErrchk( cudaMemcpy(&dev_result, partialSum, sizeof(double), cudaMemcpyDeviceToHost) );
printf( "n Total Sum cuda %f n",  dev_result);

这假定不完整代码中的其他内容都是正确的。我把找出这三种变体之间的区别留给读者作为练习。

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