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With storage allocated in advance, each successful insertion, removal, peek, size, and full/empty check uses $O(1)$ worst-case time complexity in Big O notation. This assumes constant-cost indexed array access, bounded machine word index arithmetic and fixed-cost scalar item reads and writes. Each operation touches a bounded number of slots and fields regardless of $n$ and $C$; no shifting, scanning or resizing occurs. Rejected operations also take constant work. Producing a copy of all $n$ live items takes $O(n)$ work under the same scalar-cost assumptions.

The preallocated array occupies $O(C)$ scalar slots plus constant bookkeeping. Allocation and any initial slot construction are separate setup costs. Copying a large item, running an expensive destructor or callback, allocating an item, or growing a dynamic array changes the operation cost and must be charged separately. The bound is worst-case structural work, with no need for amortized analysis, and says nothing about a measured wall-clock deadline.

These formulas describe sequential calls. Wrapping array indices implies neither atomic operations nor safe access from multiple software threads. Publication of an item's contents and reuse of its slot require a protocol justified under the relevant memory model, including any necessary acquire-release synchronization. The kernel documentation discusses barriers under explicit producer/consumer restrictions; Boost's documentation separately requires mutual exclusion for shared access involving writes. Neither property follows merely from choosing a circular buffer.^[ref][ref]

A dynamic array is an array whose backing storage can grow as entries are appended. Dynamic array doubling is one growth policy with a deterministic amortized analysis.

Consider a dynamic array holding fixed-size entries. Its size $n$ is the number of occupied slots and its capacity $C$ is the number of allocated slots. Start with $n=0$, $C=1$. The only operation is append: if $n=C$, allocate a new array of capacity $2C$, copy the $C$ existing entries, and release the old array; then write the new entry and increase $n$ by one. There is no reserve operation, deletion or shrinking in this model.

Count one unit for writing the new entry and one unit per existing entry copied during resizing. Thus the $i$th append has actual cost $c_i=1$ without resizing and $c_i=C+1$ when the old capacity $C$ is full. Copying an entry, including its read and write, is one counted action. Allocation, deallocation, index arithmetic operations and the initial one-slot allocation are outside this exact count. If these add only a constant per ordinary append and a linear cost per resize, the Big O notation conclusions still hold, with different constants; costly entry copies require their own cost model.

The first ten costs are $1,2,3,1,5,1,1,1,9,1$. For example, append $9$ copies eight entries and writes one, so $c_9=9$. At that point the capacity becomes $16$, but the total cost of all nine appends is only $24$.

For $n\ge 1$, the final capacity is $C_n=2^{\lceil {\log }_{2}n\rceil }$. Every earlier capacity $1,2,4,\dots ,C_n/2$ was copied once. This geometric series sums to $C_n-1$, including the empty sum when $n=1$. Therefore aggregate analysis gives the exact totalConsequently $T(n)<3n$, so append has amortized $O(1)$ cost in Big O notation for every prefix from this empty initial state. The values being appended do not affect the count. In contrast, an append to a full dynamic array with $n$ entries costs $n+1$: its worst-case cost is $O(n)$, and actually grows linearly along full-capacity states. A bound on total work does not promise a constant response time for each call.

The same policy controls computer memory: after each completed append, $n\le C_n<2n$ for $n\ge 1$, so fewer than $n$ slots are unused. This is a bound on the retained backing array, with fixed-size slots. During a resize from full capacity $k$ to $2k$, a copying implementation can temporarily hold both arrays, totalling $3k$ allocated slots before releasing the old one. Thus the retained-capacity bound does not bound this temporary peak by twice the live size. Allocator metadata and separately allocated objects are additional computer memory costs.

The geometric series is the source of the efficiency. If capacity instead increases by one each time, the same count becomes $n+1+2+\cdots +(n-1)=n(n+1)/2$, a quadratic total. Dynamic array doubling buys a linear total with bounded spare capacity, even when the final length is unknown.