I'm trying understand what 0xFF does under the hood in the following code snippet:
if cv2.waitKey(0) & 0xFF == ord('q'): break
Any ideas?
Answer #1: 0xFF is a hexadecimal constant which is 11111111 in binary. By using bitwise AND ( & ) with this constant, it leaves only the last 8 bits of the original (in this case, whatever cv2. waitKey(0) is). Answered By: Kevin W.
Python OpenCV – waitKey() Function waitkey() function of Python OpenCV allows users to display a window for given milliseconds or until any key is pressed. It takes time in milliseconds as a parameter and waits for the given time to destroy the window, if 0 is passed in the argument it waits till any key is pressed.
4. cv2. waitKey() This function is very important, without this function cv2.
It is also important to note that ord('q') can return different numbers if you have NumLock activated (maybe it is also happening with other keys). For example, when pressing c, the code:
key = cv2.waitKey(10) print(key)
returns
1048675 when NumLock is activated 99 otherwise
Converting these 2 numbers to binary we can see:
1048675 = 100000000000001100011 99 = 1100011
As we can see, the last byte is identical. Then it is necessary to take just this last byte as the rest is caused because of the state of NumLock. Thus, we perform:
key = cv2.waitKey(33) & 0b11111111 # 0b11111111 is equivalent to 0xFF
and the value of key will remain the same and now we can compare it with any key we would like such as your question
if key == ord('q'):
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