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Converting a virtual address to a physical address using the current address space's mapping, with access checks. In paging, the memory page number changes while the offset within it stays fixed.

Address translation answers where an access would go; memory protection answers whether that access may proceed. Neither answer is the stored value. Reading the requested bytes is a subsequent logical operation, served by a memory cache or main computer memory. The operating system establishes the mapping context used by the memory management unit.^[ref]

Consider an original toy model with unsigned 12-bit virtual addresses, byte addressing, and 64-byte memory pages and page frames. Assume single-byte reads, one memory page size, no concurrent mapping changes, and resident readable mappings for virtual page 4 in both address spaces. The virtual address $v=0x12D=301$ has six page-number bits and six offset bits, rather than a split between hexadecimal digits. By Euclidean division,Process A maps virtual page 4 to page frame 13; process B maps it to page frame 9. Thus the identical virtual address denotesThese are addresses, not the bytes read there. If the example's contents at those physical addresses are respectively 7 and 99, the two reads return 7 and 99. The mapping calculation alone cannot determine either value.

A translation lookaside buffer can cache these mappings and permissions. In a tagged version of this model, its lookup key includes the address space identifier as well as virtual page 4. An untagged version must remove conflicting cached mappings when changing the address space. Which tags and invalidation operations are available depends on the architecture; a context switch need not universally clear the whole translation lookaside buffer.^[ref]

The arithmetic explains the beauty of paging: a mapping can move an entire memory page without changing its internal positions. Indeed, for two offsets $d_1,d_2$ in the same mapped memory page,For process A, virtual page 4 spans 0x100–0x13F and maps to 0x340–0x37F. Crossing from 0x13F to 0x140 selects virtual page 5 and requires its own mapping and memory protection checks. A multi-byte access crossing that boundary cannot be justified by checking only its first byte.

Paging divides an address space into fixed-size memory pages and main computer memory into equally sized page frames. Here assume byte addressing, one memory page size $P=2^b$ bytes, and an unsigned virtual address $v$. Its virtual page number $q$ and offset $d$ areThe brackets denote the floor function, and $\mathrm{mod}$ gives the remainder in Euclidean division. If the page table maps $q$ to physical frame number $f$ and permits the access, the physical address isThe low $b$ bits are unchanged; the other bits select the memory page or page frame. A bit shift and bit mask can extract these fields. ^[ref]

For a 16-bit virtual address and $P=256=2^8$ bytes, take $v=0x1234=4660$ in hexadecimal. Then $q=18=0x12$ and $d=52=0x34$. A mapping to $f=7=0x07$ givesThus the eight high bits choose a mapping; the eight low bits choose the same position in the resulting page frame. The 256-byte size is for illustration, not a claim about an actual architecture's supported memory page sizes.

The useful freedom is that adjacent memory pages can occupy unrelated page frames. In the diagram, virtual pages 0x12 and 0x13 occupy page frames 0x07 and 0x02; virtual addresses 0x12ff and 0x1300 are consecutive, but translate to physical addresses 0x07ff and 0x0200. Uniform units simplify placement and page replacement, at the cost of page table storage and unused bytes within partly filled memory pages. For example, placing 300 bytes in exclusive 256-byte page frames requires two page frames, leaving $512-300=212$ bytes unused.

Extract one four-bit digit at a time, beginning at the most significant end of the Java int. A lookup string supplies characters, but no library performs the conversion:The logical right bit shift >>> inserts zeros, and the bit mask 15 retains exactly the low four bits of the shifted word. Each emitted character is therefore the corresponding hexadecimal digit. The started flag suppresses leading zero digits; the shift == 0 exception makes zero produce "0", rather than an empty string.

A negative Java integer uses two's complement. Its top bit is set, so the first extracted digit is between eight and fifteen and all eight digits are emitted. We interpret its original 32-bit pattern, without taking an absolute value or negating a potentially unnegatable minimum integer. In particular, toHex(19) returns "13", toHex(-1) returns "ffffffff", and toHex(Integer.MIN_VALUE) returns "80000000". There are exactly eight iterations, so for this fixed word size the time complexity is constant.

Choose an original teaching model with 16-bit virtual addresses, 16-bit physical addresses, and 256-byte memory pages and page frames. The eight high bits identify a virtual memory page; divide these into two four-bit indices. The eight low bits are the offset within the memory page. Thus each page table has 16 entries: the root selects a lower-level page table, and that table selects a data page frame. Every page table entry is four bytes, including its reference and flags. These are chosen model parameters, not a hardware entry encoding.

For a virtual address $v$, extract the root index $i$, lower-level index $j$, and offset $d$ by Euclidean division:Here $\lfloor \cdot \rfloor$ is the floor function and $\mathrm{mod}$ denotes the remainder in Euclidean division. Each index ranges from 0 to 15, and $d$ ranges from 0 to 255. In hexadecimal, virtual address 0x1234 splits into $i=1$, $j=2$, and $d=0x34=52$; its binary numeral system representation is 0001 | 0010 | 00110100.

Fix the requested operation to be a one-byte user read. Assume both table nodes and the data page frame are resident, table reads are allowed, and no mapping changes during the page table walk. The model requires presence at both levels and takes the set intersection of the two entries' read, write, execute, and user-access permissions. This is an explicit memory protection rule for this model; Sv39 page tables use different rules. The root's physical address is 0x8000; the relevant entries are:

The first lookup obtains a table reference, not the final data physical address. The second lookup obtains page frame number 7. Both entries permit a user read, so the original offset is added to the data page frame base:The page table walk therefore reads records at 0x8004 and 0x9008, then the requested read accesses data at 0x0734. All three are distinct physical addresses. These are two dependent table lookups followed by one data access, not a claim of three accesses to main computer memory: a memory cache can satisfy table lookups, and cached intermediate translations can skip them.

Under the same memory protection rule, a user write at 0x1234 fails because lower-level entry 2 denies writes, even though root entry 1 allows them. Clearing either entry's presence flag also prevents the access. An absent root entry stops the page table walk before any lower-level lookup. A page fault then requires the operating system to determine whether a permitted mapping needs storage or the access is invalid; absence alone does not establish demand paging eligibility.

With an empty translation lookaside buffer, the permitted read causes a TLB miss and this successful page table walk, with no page fault. If the resulting mapping and permissions are cached for the current address space, a later read at 0x1235 can use a TLB hit to reach 0x0735. A later write can cause a protection page fault on a TLB hit because the cached mapping remains read-only. Assume cached permissions are current; changing them requires the architecture's prescribed invalidation. A TLB miss describes a missing cached translation, while a page fault describes an access that cannot proceed under the translation or protection rules. ^[ref]