 Circular buffer full and empty states

With front-plus-count storage, empty means $n=0$ and full means $n=C$; every allocated slot is usable. With only the two wrapped array indices $h,w$, the equality $h=w$ describes both possibilities. Indeed $w-h\equiv n(\mathrm{mod}C)$, and the two allowed counts $0,C$ have the same remainder. Recovering the count as $(w-h)\mathrm{mod}C$ therefore loses the full state. A count, a separate full flag, or a different index representation must resolve that ambiguity.

Another convention reserves one slot: for a physical array of length $C\ge 2$, empty means $w=h$ and full means $(w+1)\mathrm{mod}C=h$. Its usable capacity is $C-1$, and $(w-h)\mathrm{mod}C$ now recovers its count because $0\le n<C$. A five-slot array then stores at most four items.

This is the convention of the kernel's documented CIRC_* helpers, which additionally require power-of-two array lengths. It is not a universal property of circular buffers; our capacity-five count-based example uses all five slots and does not use those helpers.^[ref]

The smallest count-based case, $C=1$, makes the issue vivid: both $h$ and $w$ are always zero, but $n$ still distinguishes one live item from none. Capacity zero is excluded from our formulas because remainder modulo zero is undefined.