 Aligned free-block capacity

Aligned free-block capacity is the usable payload that a particular free extent can supply after meeting memory alignment and metadata requirements. It differs from the raw extent length and from the largest free memory block. For an illustrative block $[a,b)$, header requirement $h$ and payload alignment $A>0$, the first permissible payload start isIf the allocator reserves no trailer or additional rounding, usable payload capacity is $\max (b-p,0)$. This formula belongs to the stated layout; it is not a universal allocator ABI.

For a free interval $[3,67)$ in byte offsets, an8-byte header and16-byte alignment give $p=16$. The raw64-byte extent then has51 bytes after the aligned start. Allocation alignment padding is $p-(a+h)=5$ bytes, in addition to the8-byte header. A48-byte payload fits this model, whereas a52-byte payload does not.

If allocated payload size must also be rounded to a multiple $q$, a requested size $R$ needs $q\lceil R/q\rceil$ usable bytes. For $q=8$ in the same example, the largest such payload is48 bytes: a49-byte request rounds to56 and fails. Header information and allocation granularity can therefore change the answer even when the alignment condition itself is satisfied.^[ref]

For integer addresses, the padding before the payload lies between0 and $A-1$ bytes. Splitting a chosen block can introduce another constraint: a leftover interval may be too small to hold a free record. A capacity calculation must consequently identify the actual header, alignment, rounding and remainder conventions rather than treating every raw free byte as payload.