问题

I'm implementing a CUDA program for transposing an image. I created 2 kernels. The first kernel does out of place transposition and works perfectly for any image size.

Then I created a kernel for in-place transposition of square images. However, the output is incorrect. The lower triangle of the image is transposed but the upper triangle remains the same. The resulting image has a stairs like pattern in the diagonal and the size of each step of the stairs is equal to the 2D block size which I used for my kernel.

Out-of-Place Kernel:

Works perfectly for any image size if src and dst are different.

template<typename T, int blockSize>
__global__ void kernel_transpose(T* src, T* dst, int width, int height, int srcPitch, int dstPitch)
{
    __shared__ T block[blockSize][blockSize];

    int col = blockIdx.x * blockSize + threadIdx.x;
    int row = blockIdx.y * blockSize + threadIdx.y;

    if((col < width) && (row < height))
    {
        int tid_in = row * srcPitch + col;
        block[threadIdx.y][threadIdx.x] = src[tid_in];
    }

    __syncthreads();

    col = blockIdx.y * blockSize + threadIdx.x;
    row = blockIdx.x * blockSize + threadIdx.y;

    if((col < height) && (row < width))
    {
        int tid_out = row * dstPitch + col;
        dst[tid_out] = block[threadIdx.x][threadIdx.y];
    }
}

In-Place Kernel:

template<typename T, int blockSize>
__global__ void kernel_transpose_inplace(T* srcDst, int width, int pitch)
{
    __shared__ T block[blockSize][blockSize];

    int col = blockIdx.x * blockDim.x + threadIdx.x;
    int row = blockIdx.y * blockDim.y + threadIdx.y;

    int tid_in = row * pitch + col;
    int tid_out = col * pitch + row;

    if((row < width) && (col < width))
        block[threadIdx.x][threadIdx.y] = srcDst[tid_in];

    __threadfence();

    if((row < width) && (col < width))
        srcDst[tid_out] = block[threadIdx.x][threadIdx.y];
}

Wrapper Function:

int transpose_8u_c1(unsigned char* pSrcDst, int width,int pitch)
{
    //pSrcDst is allocated using cudaMallocPitch

    dim3 block(16,16);
    dim3 grid;
    grid.x = (width + block.x - 1)/block.x;
    grid.y = (width + block.y - 1)/block.y;

    kernel_transpose_inplace<unsigned char,16><<<grid,block>>>(pSrcDst,width,pitch);

    assert(cudaSuccess == cudaDeviceSynchronize());

    return 1;
}

Sample Input & Wrong Output:

I know this problem has something to do with the logic of in-place transpose. This is because my out of place transpose kernel which is working perfectly for different source and destination, also gives the same wrong result if I pass it a single pointer for source and destination.

What am I doing wrong? Help me in correcting the In-place kernel.

回答1:

Your in-place kernel is overwriting data in the image that will be subsequently picked up by another thread to use for its transpose operation. So for a square image, you should buffer the destination data before overwriting it, then place the destination data in it's proper transposed location. Since we're doing effectively 2 copies per thread using this method, there's only a need to use half as many threads. Something like this should work:

template<typename T, int blockSize>
__global__ void kernel_transpose_inplace(T* srcDst, int width, int pitch)
{

    int col = blockIdx.x * blockDim.x + threadIdx.x;
    int row = blockIdx.y * blockDim.y + threadIdx.y;

    int tid_in = row * pitch + col;
    int tid_out = col * pitch + row;

    if((row < width) && (col < width) && (row<col)) {

        T temp = srcDst[tid_out];

        srcDst[tid_out] = srcDst[tid_in];
        srcDst[tid_in] = temp;
        }
}

来源：https://stackoverflow.com/questions/14174876/cuda-in-place-transpose-error

标签

cuda