CUDA仅处理OpenCV 16位灰度Mat中总列的一半

时间:2019-03-08 09:05:13

标签: opencv cuda 16-bit

我正在制作一个初学者CUDA程序,该程序基本上使用OpenCV对输入的灰度图像进行降采样。经过测试,它可以很好地处理8位灰度图像,但是当输入16位灰度图像时,它会给出嘈杂的降采样图像,图像的右半部分为空白。以下是我编写的代码。

提供了示例输入和输出图像

enter image description here

enter image description here

我的main.cpp代码将图像加载到Mat中:

int main()
{
    cv::Mat im1 = cv::imread("test.png", -1);
    std::string output_file = "resultout.png";
    binFilter(im1, output_file);

    return 0;
}

我的CUDA内核代码:

__global__ void binCUDAKernel(unsigned char *input, unsigned char *output, int binDim, int outputWidth, int outputHeight, int inputWstep, int outputWstep, int nChannels)
    {
        int outXind = blockIdx.x * blockDim.x + threadIdx.x;
        int outYind = blockIdx.y * blockDim.y + threadIdx.y;
        if ((outXind < outputWidth) && (outYind < outputHeight)) // Only run threads in output image coordinate range
        {
            if (nChannels == 1) // Test only for greyscale images
            {
                // Calculate x & y index of input binned pixels corresponding to current output pixel
                int inXstart = outXind * binDim;
                int inYstart = outYind * binDim;

                // Perform binning on identified input pixels
                float sum = 0;
                for (int binY = inYstart; binY < (inYstart + binDim); binY++) {
                    for (int binX = inXstart; binX < (inXstart + binDim); binX++) {
                        int input_tid = binY * inputWstep + binX;
                        sum += input[input_tid];
                    }
                }

                // Establish output thread index in current output pixel index
                int output_tid = outYind * outputWstep + outXind;

                // Assign binned pixel value to output pixel
                output[output_tid] = static_cast<unsigned short>(sum / (binDim*binDim));
            }
        }
    }

我的CPU代码:

void binFilter(const cv::Mat input, std::string output_file)
{
    // 2X2 binning
    int binDim = 2;

    // Create blank output image & calculate size of input and output
    cv::Size outsize(input.size().width / binDim, input.size().height / binDim);
    cv::Mat output(outsize, input.type());
    const int inputBytes = input.step * input.rows;
    const int outputBytes = output.step * output.rows;

    // Allocate memory in device
    unsigned char *d_input, *d_output;
    gpuErrchk(cudaMalloc<unsigned char>(&d_input, inputBytes));
    gpuErrchk(cudaMalloc<unsigned char>(&d_output, outputBytes));

    // Copy input image to device
    gpuErrchk(cudaMemcpy(d_input, input.ptr(), inputBytes, cudaMemcpyHostToDevice));

    // Configure size of block and grid
    const dim3 block(16, 16);
    const dim3 grid((output.cols + block.x - 1) / block.x, (output.rows + block.y - 1) / block.y); // Additional block for rounding up

    // Execute kernel
    binCUDAKernel <<<grid, block>>> (d_input, d_output, binDim, output.cols, output.rows, input.step, output.step, input.channels());
    gpuErrchk(cudaPeekAtLastError());

    // Wait for all threads to finish
    //gpuErrchk(cudaDeviceSynchronize());

    // Copy output image from device back to host (cudaMemcpy is a blocking instruction)
    gpuErrchk(cudaMemcpy(output.ptr(), d_output, outputBytes, cudaMemcpyDeviceToHost));

    // Free device memory
    gpuErrchk(cudaFree(d_input));
    gpuErrchk(cudaFree(d_output));

    // Write image to specified output_file path
    cv::imwrite(output_file, output);
}

我怀疑这可能是某种数据类型不匹配,但我无法弄清楚。

1 个答案:

答案 0 :(得分:2)

首先,为了处理16位图像,必须将像素数据解释为16位宽的数据类型,该数据类型可能为unsigned shortshort。请记住,我们只需要将图像数据解释为unsigned short类型;不用打字为此,我们仅将图像数据指针转换为所需的类型,如以下示例所示:

unsigned short* ptr16 = reinterpret_cast<unsigned short*>(im1.ptr());

由于上述步骤,我们还必须为16位数据类型创建一个单独的内核。我们可以通过将内核定义为C ++模板来巧妙地做到这一点。 因此内核可能如下所示:

template<typename T>
__global__ void binCUDAKernel(T *input, T *output, int binDim, int outputWidth, int outputHeight, int inputWstep, int outputWstep, int nChannels)
{
    int outXind = blockIdx.x * blockDim.x + threadIdx.x;
    int outYind = blockIdx.y * blockDim.y + threadIdx.y;

    if ((outXind < outputWidth) && (outXind > outputWidth/2) && (outYind < outputHeight)) // Only run threads in output image coordinate range
    {
        if (nChannels == 1) // Test only for greyscale images
        {
            // Calculate x & y index of input binned pixels corresponding to current output pixel
            int inXstart = outXind * binDim;
            int inYstart = outYind * binDim;

            // Perform binning on identified input pixels
            float sum = 0;
            for (int binY = inYstart; binY < (inYstart + binDim); binY++) {
                for (int binX = inXstart; binX < (inXstart + binDim); binX++) {
                    int input_tid = binY * inputWstep + binX;
                    sum += float(input[input_tid]);
                }
            }

            // Establish output thread index in current output pixel index
            int output_tid = outYind * outputWstep + outXind;

            // Assign binned pixel value to output pixel
            output[output_tid] = static_cast<T>(sum / (binDim*binDim));
        }
    }
}   

在使用自定义CUDA内核处理OpenCV Mat的过程中,另一个重要的问题是,图像步长必须除以数据类型的大小(以字节为单位)。对于16位图像,单个像素的大小为16位(2字节),因此内核中使用的步骤必须除以2。请记住,原始步骤不应修改。应该只对作为内核参数传递的步长值进行除法。

结合以上修复程序,最终的CPU代码可能如下所示:

void binFilter(const cv::Mat input, std::string output_file)
{
    // 2X2 binning
    int binDim = 2;

    // Create blank output image & calculate size of input and output
    cv::Size outsize(input.size().width / binDim, input.size().height / binDim);
    cv::Mat output(outsize, input.type());
    const int inputBytes = input.step * input.rows;
    const int outputBytes = output.step * output.rows;

    // Allocate memory in device
    unsigned char *d_input, *d_output;
    gpuErrchk(cudaMalloc<unsigned char>(&d_input, inputBytes));
    gpuErrchk(cudaMalloc<unsigned char>(&d_output, outputBytes));

    // Copy input image to device
    gpuErrchk(cudaMemcpy(d_input, input.ptr(), inputBytes, cudaMemcpyHostToDevice));

    // Configure size of block and grid
    const dim3 block(16, 16);
    const dim3 grid((output.cols + block.x - 1) / block.x, (output.rows + block.y - 1) / block.y); // Additional block for rounding up


    int depth = input.depth();
    // Execute kernel

    if (input.depth() == CV_16U)
    {
        typedef unsigned short t16;

        t16* input16 = reinterpret_cast<t16*>(d_input);
        t16* output16 = reinterpret_cast<t16*>(d_output);

        int inputStep16 = input.step / sizeof(t16);
        int outputStep16 = output.step / sizeof(t16);

        binCUDAKernel <t16> <<<grid, block>>> (input16, output16, binDim, output.cols, output.rows, inputStep16, outputStep16, input.channels());
    }
    else
    {
        binCUDAKernel <unsigned char> <<<grid, block>>> (d_input, d_output, binDim, output.cols, output.rows, input.step, output.step, input.channels());   
    }


    gpuErrchk(cudaPeekAtLastError());

    // Wait for all threads to finish
    //gpuErrchk(cudaDeviceSynchronize());

    // Copy output image from device back to host (cudaMemcpy is a blocking instruction)
    gpuErrchk(cudaMemcpy(output.ptr(), d_output, outputBytes, cudaMemcpyDeviceToHost));

    // Free device memory
    gpuErrchk(cudaFree(d_input));
    gpuErrchk(cudaFree(d_output));

    // Write image to specified output_file path
    cv::imwrite(output_file, output);
}

由于合并算法的逻辑,输出图像中的噪声似乎引入了混叠。例如,它与使用最近邻居方法对图像重新采样非常相似。

更新:

上面提到的计算像素的内存地址的方法是没有记载的,只是直觉的结果,因此似乎有点不合常规。 OpenCV和其他库使用的另一种方法避免了图像步骤划分的混乱。给定一个像素的x和y索引,操作如下:

  1. 将图像数据指针重新解释为字节表示形式(unsigned char*)。
  2. 使用y索引和图像步长计算图像行的起始地址。
  3. 将行起始地址重新解释为必需的类型(unsigned short*)。
  4. 访问行起始指针的x索引。

使用此方法,我们可以如下计算灰度图像的像素存储地址:

template<typename T>
T* getPixelAddress(unsigned char* data, int x, int y, int step)
{
    T* row = (T*)((unsigned char*)(data) + y * step);
    return row + x;
}

在上述方法中,步长值是原始值,没有任何除法。