What's 0xFF for in cv2.waitKey(1)?

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?

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'):

cv2.waitKey() returns a 32 Bit integer value (might be dependent on the platform). The key input is in ASCII which is an 8 Bit integer value. So you only care about these 8 bits and want all other bits to be 0. This you can achieve with:

cv2.waitKey(0) & 0xFF

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).