On Tue, 26 Oct 2021 22:21:23 -0400 Steven Rostedt <rostedt@xxxxxxxxxxx> wrote: > I'm sure there's an algorithm somewhere that can give as the real max. You got me playing with this more ;-) OK, I added the rounding in the wrong place. I found that we can make the max_div to be the same as the shift! The bigger the shift, the bigger the max! mult = (1 << shift) / div; max_div = (1 << shift) But the rounding needs to be with the mult / shift: return (val * mult + ((1 << shift) - 1)) >> shift; When val goes pass 1 << shift, then the error will be off by more than one. -- Steve
