Question

点击我之前的帖子：CUDA NPP - unknown error upon GPU error check

我试图通过使用CUDA NPP库来总结图像中的所有像素，并且在一些开发人员的帮助下，我终于得到了我的代码来编译。但是，当我尝试打印出partialSum中存储的值时，将其复制到double变量中（与CUDA v4.2的NPP指南一致），我收到此错误：

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

我一直在试图摆脱它，但到目前为止我还没有成功。请帮忙！我现在有两天的代码。

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 %d\n", 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

Answer 1

这里的问题似乎是基本的指针滥用（我说似乎是因为我们有不完整的，无法编译的代码，所以很难肯定地说）。

这应该有效：

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

这也应该有效：

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所有内容都是正确的。我把它作为练习留给读者来发现三种变体之间的差异。

CUDA NPP - 打印输出错误

1 个答案: