# rank question **URL:** \ **Category:** Community Support\ **Tags:** kdb-and-q\ **Created:** [December 14, 2015, 2:04am UTC](https://forum.kx.com/t/rank-question/10681 "2015-12-14T02:04:00Z")\ **Posts on this page:** 1\ **Page:** 1
**Author:** ![User7](https://avatars.discourse-cdn.com/v4/letter/u/7ea924/32.png) [@User7](https://forum.kx.com/u/User7)\ **Post date:** [December 14, 2015, 2:04am UTC](https://forum.kx.com/t/rank-question/10681/1 "2015-12-14T02:04:00Z")
1. Isn’t rank of an atom 0, a simple list 1 and “2 3 #til 6” 2.  2. in x?y the rank of y should be 1 less than x (I call it the rank rule).  3. I believe I have identified the x and y in the following statement correctly as x being “2 3” and y being “2 3#til6” because the webpage says “If there is no match the result is the count of the left argument” . Hence the output of 2 highlighted proves left side is x and right side is y q)2 3?2 3#til 6 2 2 0 1 2 2 **Given this I have following questions** A. Why is 2 3 claimed to be of 0 rank and (0 1 2;4 5) of 1 rank B. Isn’t the common argument in both expressions (rhs) supposed to be of 1 rank less  ? is rank sensitive; x?y can’t deal with mixed rank x. If rank x is n then x?y looks for objects of rank n-1. e.g. ``` 2 3?2 3#til 6 / looks for rank 0 objects(0 1 2;4 5)?2 3#til 6 / looks for rank 1 objects ```