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C# Tutorial: How To Flip an Image Horizontally


Andraz Krzisnik
C# Tutorial: How To Flip an Image...

Where Do We Start?

There are numerous ways you can flip an image horizontally. It’s pretty basic functionality in image processing. I wanted to challenge myself and not just use the preexisting line of code that takes care of this functionality entirely. So I coded entire thing myself, well, at least some level closer to the base anyway.

To mirror an image horizontally I used the “lockbits” method, where I rearranged pixel values stored in a one dimensional array. The problem we face here is that we only want to flip pixel values along horizontal axis while rows stay in the same order. So what we need to do is separate rows and columns somehow.

We Need To Track Our Pixels

Luckily, we know that a row has a length of one “Stride” value and each pixel holds 4 values (red, green, blue and alpha values). With that in mind, we will nest 3 for loops within each other.

Like this:

for ( ... ){
   for( ... ){
      for( ... ){
         some code...
         }
      }
}

Now we need to create a variable that will get the position for any pixel in the array of pixel values. So, the nested for loops we created will help us keep track of what exact value we are processing. Most inner for loop will iterate through RGB values, the middle and most outer loop will iterate through horizontal and vertical position of the pixels.

Let’s See Where The Borders Are

Now the last thing we need, are limits for each for loop to work. For RGB for loop we declare the limit to be 3, as there are values for red, green and blue color channel. This is a constant number, which means it’s applied for every image we are processing. The fourth pixel value, alpha channel, which controls the opacity of each pixel will be set to it’s maximum. We are processing 32-bit images so the maximum for each value is 255. 32 times 4 values is 256 possible values for each channel. 255 is the maximum value, because counting starts with 0.

Limits for horizontal and vertical values are the dimensions of the image we are processing. Remember not to mix Stride and Width of the image, since Stride includes each pixel values while Width of the image holds only the number of pixels in a row.

What’s The Magic Equation?

Now that we know about all the numbers we are working with, we can return to the variable that will tell us which pixel we are processing.

On the vertical or “y” coordinate we will move with “y * Stride”. “y” is the variable which we are iterating. To move along the horizontal or “x” coordinate we will use “x * 4”. “x” being the variable that is being iterated and 4, which stands for 4 channels for each pixel.

Now if we put it together, it should look something like this:

current_pixel = y * sourceData.Stride + x * 4;

The Entire Function That Will Flip an Image

private Bitmap HorizontalFlip(Bitmap img)
{
    int w = img.Width;
    int h = img.Height;
    BitmapData sd = img.LockBits(new Rectangle(0, 0, w, h),
        ImageLockMode.ReadWrite, PixelFormat.Format32bppArgb);
    int bytes = sd.Stride * sd.Height;
    byte[] buffer = new byte[bytes];
    byte[] result = new byte[bytes];
    Marshal.Copy(sd.Scan0, buffer, 0, bytes);
    img.UnlockBits(sd);
    int current, flipped = 0;
    for (int y = 0; y < h; y++)
    {
        for (int x = 4; x < w; x++)
        {
            current = y * sd.Stride + x * 4;
            flipped = y * sd.Stride + (w - x) * 4;
            for (int i = 0; i < 3; i++)
            {
                result[flipped + i] = buffer[current + i];
            }
            result[flipped + 3] = 255;
        }
    }
    Bitmap resimg = new Bitmap(w, h);
    BitmapData rd = resimg.LockBits(new Rectangle(0, 0, w, h),
        ImageLockMode.WriteOnly, PixelFormat.Format32bppArgb);
    Marshal.Copy(result, 0, rd.Scan0, bytes);
    resimg.UnlockBits(rd);
    return resimg;
}

Demonstration

flip an image

 

 

Get Complete Project Here

 

Download Project

 

 

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