How does BitmapData solve the problem of binarization of Verification Code images,

How does BitmapData solve the problem of binarization of Verification Code images,

Sorry, this is an article for help. First, it's hard to avoid being clear and zoom in a bit. below is the picture. There is no binarization above. below is the result after binarization, in fact, I don't know what binary is, that is, a certain range turns black and white.

Problem:

Why are there a lot of colored dots on my results? None of them ......

Who can help me see how to change the code! Thank you !!

Bitmap bit1 = new Bitmap (bit); Rectangle rect1 = new Rectangle (0, 0, bit1.Width, bit1.Height); BitmapData bitd = bit1.LockBits (rect1, ImageLockMode. readWrite, PixelFormat. format24bppRgb); IntPtr srcPtr = bitd. scan0; int scanWidth = bitd. width * 3; int src_bytes = scanWidth * bitd. height; byte [] srcRGBValues = new byte [src_bytes]; Marshal. copy (srcPtr, srcRGBValues, 0, src_bytes); // grayscale processing int m = 0, I = 0, j = 0; // m indicates the row, j indicates that the column int k = 0; for (I = 0; I <bit. height; I ++) // only obtain the rows row pixel value of the image {for (j = 0; j <bit. width; j ++) {// only process the pixel data of each row of the image and Discard unused space. // note that in the bitmap structure, RGB is stored in BGR order. k = 3 * j; var grays = Convert. toInt16 (srcRGBValues [I * scanWidth + k + 2]) + Convert. toInt16 (srcRGBValues [I * scanWidth + k + 1]) + Convert. toInt16 (srcRGBValues [I * scanWidth + k + 0]); if (grays >=382) {srcRGBValues [I * scanWidth + k + 2] = Convert. toByte (255); srcRGBValues [I * scanWidth + k + 1] = Convert. toByte (255); srcRGBValues [I * scanWidth + k + 0] = Convert. toByte (255);} else {srcRGBValues [I * scanWidth + k + 2] = Convert. toByte (0); srcRGBValues [I * scanWidth + k + 1] = Convert. toByte (0); srcRGBValues [I * scanWidth + k + 0] = Convert. toByte (0) ;}} m ++;} Marshal. copy (srcRGBValues, 0, srcPtr, src_bytes); bit1.UnlockBits (bitd); Image image1 = bit1;