Detect black pixel in image iOS

Question

As of now I am searching every pixel 1 by 1 checking the color and seeing if it\'s black... if it isn\'t I move on to the next pixel. This is taking forever as I can only ch

Answer 1

Why are you using a timer at all? Why not just have a double for loop in your function that loops over all possible x- and y-coordinates in the image? Surely that would be waaaay faster than only checking at most 100 pixels per second. You would want to have the x (width) coordinates in the outer loop and the y (height) coordinates in the inner loop so that you are effectively scanning one column of pixels at a time from left to right, since you are trying to find the leftmost black pixel.

Also, are you sure that each pixel in your image has a 4-byte (Uint32) representation? A standard bitmap would have 3 bytes per pixel. To check if a pixel is close to black, you would just examine each byte in the pixel separately and make sure they are all less than some threshold.

EDIT: OK, since you are using UIGetScreenImage, I'm going to assume that it is 4-bytes per pixel.

const UInt8 *pixels = CFDataGetBytePtr(imageData);
UInt8 blackThreshold = 10; // or some value close to 0
int bytesPerPixel = 4;
for(int x = 0; x < width1; x++) {
  for(int y = 0; y < height1; y++) {
    int pixelStartIndex = (x + (y * width1)) * bytesPerPixel;
    UInt8 alphaVal = pixels[pixelStartIndex]; // can probably ignore this value
    UInt8 redVal = pixels[pixelStartIndex + 1];
    UInt8 greenVal = pixels[pixelStartIndex + 2];
    UInt8 blueVal = pixels[pixelStartIndex + 3];
    if(redVal < blackThreshold && blueVal < blackThreshold && greenVal < blackThreshold) {
      //This pixel is close to black...do something with it
    }
  }
}

If it turns out that bytesPerPixel is 3, then change that value accordingly, remove the alphaVal from the for loop, and subtract 1 from the indices of the red, green, and blue values.

Also, my current understanding is that UIGetScreenImage is considered a private function that Apple may or may not reject you for using.