Sobel filter in cuda (cant show full image) - image

Question

I have a classic problem about the output of sobel filter using CUDA.
this is a main class (main.cpp)
/*main class */
int main(int argc, char** argv)
{
IplImage* image_source = cvLoadImage("test.jpg",
CV_LOAD_IMAGE_GRAYSCALE);
IplImage* image_input = cvCreateImage(cvGetSize(image_source),
IPL_DEPTH_8U,image_source->nChannels);
IplImage* image_output = cvCreateImage(cvGetSize(image_source),
IPL_DEPTH_8U,image_source->nChannels);
/* Convert from IplImage tofloat */
cvConvert(image_source,image_input);
unsigned char *h_out = (unsigned char*)image_output->imageData;
unsigned char *h_in = (unsigned char*)image_input->imageData;
width = image_input->width;
height = image_input->height;
widthStep = image_input->widthStep;
sobel_parallel(h_in, h_out, width, height, widthStep);
cvShowImage( "CPU", image_output );
cvReleaseImage( &image_output );
waitKey(0);
}
And this is the CUDA file (kernel_gpu.cu)
__global__ void kernel ( unsigned char *d_in , unsigned char *d_out , int width ,
int height, int widthStep ) {
int col = blockIdx . x * blockDim . x + threadIdx . x ;
int row = blockIdx . y * blockDim . y + threadIdx . y ;
int dx [3][3] = { -1 , 0 , 1 ,
-2 , 0 , 2 ,
-1 , 0 , 1};
int dy [3][3] = {1 ,2 ,1 ,
0 ,0 ,0 ,
-1 , -2 , -1};
int s;
if( col < width && row < height)
{
int i = row;
int j = col;
// apply kernel in X direction
int sum_x=0;
for(int m=-1; m<=1; m++)
for(int n=-1; n<=1; n++)
{
s=d_in[(i+m)*widthStep+j+n]; // get the (i,j) pixel value
sum_x+=s*dx[m+1][n+1];
}
// apply kernel in Y direction
int sum_y=0;
for(int m=-1; m<=1; m++)
for(int n=-1; n<=1; n++)
{
s=d_in[(i+m)*widthStep+j+n]; // get the (i,j) pixel value
sum_y+=s*dy[m+1][n+1];
}
int sum=abs(sum_x)+abs(sum_y);
if (sum>255)
sum=255;
d_out[i*widthStep+j]=sum; // set the (i,j) pixel value
}
}
// Kernel Calling Function
extern "C" void sobel_parallel( unsigned char* h_in, unsigned char* h_out,
int rows, int cols, int widthStep){
unsigned char* d_in;
unsigned char* d_out;
cudaMalloc((void**) &d_in, rows*cols);
cudaMalloc((void**) &d_out, rows*cols);
cudaMemcpy(d_in, h_in, rows*cols*sizeof( unsigned char), cudaMemcpyHostToDevice);
dim3 block (16,16);
dim3 grid ((rows * cols) / 256.0);
kernel<<<grid,block>>>(d_in, d_out, rows, cols, widthStep);
cudaMemcpy(h_out, d_out, rows*cols*sizeof( unsigned char), cudaMemcpyDeviceToHost);
cudaFree(d_in);
cudaFree(d_out);
}
Error :
the result image does not appear in their entirety, only part of the image.
Why is the result(GPU) like this?? (I tried to make CPU computation using the same function and no problem).

Answer 1

You are creating 1 Dimensional grid, while using 2D indexing inside the kernel which will cover only the x direction and only the top 16 rows of the image will be filtered (because the height of the block is 16).
dim3 grid ((rows * cols) / 256.0); //This is incorrect in current case
Consider creating 2 dimensional grid, so that it spans all the rows of the image.
dim3 grid ((cols + 15)/16, (rows + 15)/16);

Answer 2

Check the width and widthStep variables to see if they are actually equal or not because in your sobel_parallel function you are implicitly assuming this (which might not be true since your data is aligned). If this is not true the code
cudaMalloc((void**) &d_in, rows*cols);
will actually allocate less memory than necessary and hence you will only process part of your image. It would be better to use
cudaMalloc((void**) &d_in, rows*widthStep);
And of course adjust the rest of your code as necessary.
You are also calling
void sobel_parallel( unsigned char* h_in, unsigned char* h_out,
int rows, int cols, int widthStep)
with
sobel_parallel(h_in, h_out, width, height, widthStep);
which exchanges rows with cols and this is again exchanged when you are calling your kernel. This will cause a problem when you use the above suggestion.