Dilation Image on OpenCV C/C++

Question

I want to create a dilation image using a kernel that runs through the entire image and checks if the kernel zone has 0, if so, it gives the new image a pixel of 255. My code is giving me an all black dst and I don't know why. This is the code:

Mat vcpi_binary_dilate(Mat src)
{
    if (src.empty())
    {
        cout << "Failed to load image.";
        return src;
    }

    Mat dst(src.rows, src.cols, CV_8UC1, Scalar(0));

    const int kernnel = 3;
    int array[kernnel * kernnel] = {};

    for (int y = kernnel / 2; y < src.cols - kernnel / 2; y++)
    {
        for (int x = kernnel / 2; x < src.rows - kernnel / 2; x++)
            for (int yk = -kernnel / 2; yk <= kernnel / 2; yk++)
            {
                for (int xk = -kernnel / 2; xk <= kernnel / 2; xk++)
                {
                    if (src.at<uchar>(y + yk, x + xk) == 0)
                    {
                        dst.at<uchar>(y + yk, x + xk) = 255;
                    }
                }
            }
    }

    imshow("Image ", src);
    imshow("Image dilate", dst);
    waitKey(0);

    return dst;
}

I hope to have an output image of this type.

inplace dilation will produce broken results. you are modifying data you need to keep untouched for this calculation. — Christoph Rackwitz, Feb 06 '23 at 09:48
you can fundamentally not perform **binary** morphology on a **grayscale** image. all pixels will be nonzero because it's a real picture with no exactly black pixels. — Christoph Rackwitz, Feb 06 '23 at 09:50

wohlstad · Answer 1 · 2023-02-06T09:52:43.370

I am not sure about the algorithm you are trying to implement.

But there is one thing that is definately wrong:
The image dimensions are mixed up-
cols is the width and corresponds to the x axis.
rows is the height and corresponds to the y axis.
This is causing you to access the images using cv::Mat::at with invalid coordinates.

Therefore you need to change:

for (int y = kernnel / 2; y < src.cols - kernnel / 2; y++)
{
    for (int x = kernnel / 2; x < src.rows - kernnel / 2; x++)
    {

To:

//--------------------------------vvvv--------------------
for (int y = kernnel / 2; y < src.rows - kernnel / 2; y++)
{
//------------------------------------vvvv--------------------
    for (int x = kernnel / 2; x < src.cols - kernnel / 2; x++)
    {

Note that this is consistent with your calls ...at<uchar>(y + yk, x + xk), since cv::Mat::at expects the row (i.e. y coordinate) first.

A side note: Why is "using namespace std;" considered bad practice?.

Edit:

After having a look at the matlab code in your comment, which applies an algorithm different than what you described in your question, you'll need to do the following changes:

Call something equivalent to matlab's im2bw.
Update dst current pixel, not the one in the neighborhood.

Maybe something like:

cv::Mat vcpi_binary_dilate(cv::Mat src) {

    if (src.empty()) {
        std::cout << "Failed to load image.";
        return src;
    }

    cv::threshold(src, src, 127, 255, CV_THRESH_BINARY);

    cv::Mat dst(src.rows, src.cols, CV_8UC1, cv::Scalar(255));  // <-- Replacement for im2bw, might need tuning.

    const int kernnel = 3;
    int array[kernnel * kernnel] = {};

    for (int y = kernnel / 2; y < src.rows - kernnel / 2; y++)
    {
        for (int x = kernnel / 2; x < src.cols - kernnel / 2; x++)
        {
            for (int yk = -kernnel / 2; yk <= kernnel / 2; yk++)
            {
                for (int xk = -kernnel / 2; xk <= kernnel / 2; xk++)
                {
                    if (src.at<uchar>(y + yk, x + xk) == 0) {
                        dst.at<uchar>(y, x) = 0;   // <-- Update the current pixel, not the one in the neighborhood.
                    }
                }
            }
        }
    }

    cv::imshow("Image ", src);
    cv::imshow("Image dilate", dst);
    cv::waitKey(0);
    return dst;
}

I changed it as you said, it still gives me the completely black image — paulosousa32, Feb 05 '23 at 19:15
I ran your code and got white pixels as well. Could it be that your `src` does not contain any `0` pixels (because it is the only trigger for setting `dst` to 255) ? — wohlstad, Feb 05 '23 at 20:45
expect this kind of dst https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.geeksforgeeks.org%2Fmatlab-erosion-of-an-image%2F&psig=AOvVaw0pc6_GgC5WsIBK9OWtUfOu&ust=1675719466391000&source=images&cd=vfe&ved=0CA0QjRxqFwoTCODyl9Or__wCFQAAAAAdAAAAABAD — paulosousa32, Feb 05 '23 at 21:38
lenna1.png (the image there) does not contain any `0` pixel. I think your interpretation of the matlab code: "_if the kernel zone has 0, if so, it gives the new image a pixel of 255_" is incorrect. But I'm an not an expert in matlab code. — wohlstad, Feb 06 '23 at 05:11
I now see that the matlab code contains a call to `im2bw` _before_ running the kernel. You should add something similar in you c++ code. Maybe something like: `cv::threshold(src, src, 127, 255, CV_THRESH_BINARY);`, not sure. Anyway the matlab code is not equivalent. — wohlstad, Feb 06 '23 at 05:21
Looks like the matlab code is setting each target pixel to 0 if any of the neighborhood pixels are 0, and 1 otherwise. — wohlstad, Feb 06 '23 at 05:29
Also you need to load the src image as grayscsle, not color. — wohlstad, Feb 06 '23 at 06:15

Dilation Image on OpenCV C/C++

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