Finding the next in round-robin scheduling by bit twiddling

Question

Consider the following problem. You have a bit-string that represents the current scheduled slave in one-hot encoding. For example, \"00000100\" (with the leftmost bit being #7

Answer 1

Interesting problem! I can't help but wonder if you can't simplify your scheduler operation so this sort of operation would be necessary.

Given that you know VHDL, I won't go into detail, but my suggestion would be the following:

Use a 3 bit encoder to turn the currently scheduled task into a number:

01000000 --> 6

Then use a barrel shifter to rotate the mask by that number + 1 (to skip the current task):

00001010 --> 00010100

Then use a priority encoder to find the first available "next" task:

00010100 --> 00000100 --> 2

Then reverse the barrel shift by addition:

(2+7) % 8 = 1

Which when re-encoded will give the next scheduled task:

00000010

Should be very fast and straightforward, although the barrel shifter is 'expensive' in terms of realestate, but I don't see an easy way to get around that at the moment.

Edit: Doug's solution is significantly more elegant...

-Adam