Circular buffer indices ($h,n,w$)

Let the capacity be a positive integer $C$, and allocate exactly $C$ array slots $B_0,\dots ,B_{C-1}$. Store the front array index $h$ and count $n$. The representation invariant isThe logical sequence and next-write array index areHere $\mathrm{mod}$ denotes the nonnegative remainder from modular arithmetic. The count determines which slots are live: leftover bits in other slots are not members of the FIFO queue. If $n=0$, $h$ has no current front item; it identifies the position where the next successful insertion begins. If $n=C$, $w=h$ points at a live item, so writing there requires an explicit overwrite policy.

For $n<C$, enqueue(x) writes $B_w\leftarrow x$, then increments $n$, leaving $h$ fixed. For $n>0$, dequeue() saves and returns $B_h$, then sets $h\leftarrow (h+1)\mathrm{mod}C$ and $n\leftarrow n-1$. These are sequential updates, completed before the next call. A rejecting interface tests $n=C$ or $n=0$ first and returns full or empty, respectively, without changing state. peek returns $B_h$ only when $n>0$; size returns $n$.

One can instead store $w$ and $n$, recovering $h=(w-n)\mathrm{mod}C$. On insertion, advance $w$ and increment $n$; on removal, keep $w$ fixed and decrement $n$. Thus the next-write array index is not necessarily the front array index.

Names such as head and tail require checking the actual API: the kernel documentation calls the insertion position head and the removal position tail.^[ref]

The diagram is the state after step 8 of circular buffer operation trace. Reading the physical array from slot zero would give $F,B,C,D,E$, which includes a removed label and puts $F$ first. Reading from $h$ for exactly $n$ items gives $(C,D,E,F)$. The next insertion uses slot $1$, not slot $0$. This distinction also explains why the live sequence may need two contiguous pieces of the array: here slots $2$ through $4$, followed by slot $0$. No items need to move to make this wrap work.