This tutorial explains the very basics of how paging works, with focus on x86, although most high level concepts will also apply to other instruction set architectures, e.g. ARM.
The goals are to:
This tutorial was extracted and expanded from this Stack Overflow answer.
Minimal example: github.com/cirosantilli/x86-bare-metal-examples/blob/5c672f73884a487414b3e21bd9e579c67cd77621/paging.S
Like everything else in programming, the only way to really understand this is to play with minimal examples.
Paging makes it easier to compile and run two programs or threads at the same time on a single computer.
For example, when you compile two programs, the compiler does not know if they are going to be running at the same time or not.
And thread stacks, that must be contiguous and keep growing down until they overwrite each other, are an even bigger issue!
But if two programs use the same address and run at the same time, this is obviously going to break them!
Paging solves this problem beautifully by adding one degree of indirection:
(logical) ------------> (physical)
pagingWhere:
As far as programs are concerned, they think they can use any address between 0 and 4 GiB (2^32,
FFFFFFFF) on 32-bit systems.The OS then sets up paging so that identical logical addresses will go into different physical addresses and not overwrite each other.
This makes it much simpler to compile programs and run them at the same time.
Paging achieves that goal, and in addition:
- the switch between programs is very fast, because it is implemented by hardware
- the memory of both programs can grow and shrink as needed without too much fragmentation
- one program can never access the memory of another program, even if it wanted to.This is good both for security, and to prevent bugs in one program from crashing other programs.
Or if you like non-funny jokes:
Comparison between the Linux kernel userland memory virtualization and The Matrix
. Source. Is this RAM real?Paging is implemented by the CPU hardware itself.
Paging could be implemented in software, but that would be too slow, because every single RAM memory access uses it!
Operating systems must setup and control paging by communicating to the CPU hardware. This is done mostly via:
- the CR3 register, which tells the CPU where the page table is in RAM memory
- writing the correct paging data structures to the RAM pointed to the CR3 register.Using RAM data structures is a common technique when lots of data must be transmitted to the CPU as it would cost too much to have such a large CPU register.The format of the configuration data structures is fixed by the hardware, but it is up to the OS to set up and manage those data structures on RAM correctly, and to tell the hardware where to find them (via
cr3).Then some heavy caching is done to ensure that the RAM access will be fast, in particular using the TLB.Another notable example of RAM data structure used by the CPU is the IDT which sets up interrupt handlers. - CR3 cannot be modified in ring 3. The OS runs in ring 0. See also:
- the page table structures are made invisible to the process using paging itself!
Processes can however make requests to the OS that cause the page tables to be modified, notably:
- stack size changes
brkandmmapcalls, see also: stackoverflow.com/questions/6988487/what-does-brk-system-call-do/31082353#31082353
The kernel then decides if the request will be granted or not in a controlled manner.
In x86 systems, there may actually be 2 address translation steps:like this:
- first segmentation
- then paging
(logical) ------------------> (linear) ------------> (physical)
segmentation pagingThe major difference between paging and segmentation is that:
- paging splits RAM into equal sized chunks called pages
- segmentation splits memory into chunks of arbitrary sizes
This is the main advantage of paging, since equal sized chunks make things more manageable by reducing memory fragmentation problems. See also:
Paging came after segmentation historically, and largely replaced it for the implementation of virtual memory in modern OSs.
Paging has become so much more popular that support for segmentation was dropped in x86-64 in 64-bit mode, the main mode of operation for new software, where it only exists in compatibility mode, which emulates IA-32.
x86 Paging Tutorial Example: simplified single-level paging scheme by 
Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16 x86 Paging Tutorial Single level paging scheme visualization by Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16This is how the memory could look like in a single level paging scheme:
Links Data Physical address
+-----------------------+ 2^32 - 1
| |
. .
| |
+-----------------------+ page0 + 4k
| data of page 0 |
+---->+-----------------------+ page0
| | |
| . .
| | |
| +-----------------------+ pageN + 4k
| | data of page N |
| +->+-----------------------+ pageN
| | | |
| | . .
| | | |
| | +-----------------------+ CR3 + 2^20 * 4
| +--| entry[2^20-1] = pageN |
| +-----------------------+ CR3 + 2^20 - 1 * 4
| | |
| . many entires .
| | |
| +-----------------------+ CR3 + 2 * 4
| +--| entry[1] = page1 |
| | +-----------------------+ CR3 + 1 * 4
+-----| entry[0] = page0 |
| +-----------------------+ <--- CR3
| | |
| . .
| | |
| +-----------------------+ page1 + 4k
| | data of page 1 |
+->+-----------------------+ page1
| |
. .
| |
+-----------------------+ 0Notice that:
- the CR3 register points to the first entry of the page table
- the page table is just a large array with 2^20 page table entries
- each entry is 4 bytes big, so the array takes up 4 MiB
- each page table contains the physical address a page
- each page is a 4 KiB aligned 4 KiB chunk of memory that user processes may use
- we have 2^20 table entries. Since each page is 4 KiB == 2^12, this covers the whole 4 GiB (2^32) of 32-bit memory
x86 Paging Tutorial Single level paging scheme numerical translation example by Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16Suppose that the OS has setup the following page tables for process 1:and for process 2:
entry index entry address page address present
----------- ------------------ ------------ -------
0 CR3_1 + 0 * 4 0x00001 1
1 CR3_1 + 1 * 4 0x00000 1
2 CR3_1 + 2 * 4 0x00003 1
3 CR3_1 + 3 * 4 0
...
2^20-1 CR3_1 + 2^20-1 * 4 0x00005 1entry index entry address page address present
----------- ----------------- ------------ -------
0 CR3_2 + 0 * 4 0x0000A 1
1 CR3_2 + 1 * 4 0x12345 1
2 CR3_2 + 2 * 4 0
3 CR3_2 + 3 * 4 0x00003 1
...
2^20-1 CR3_2 + 2^20-1 * 4 0xFFFFF 1When process 1 tries to access a linear address, this is the physical addresses that will be actually accessed:
linear physical
--------- ---------
00000 001 00001 001
00000 002 00001 002
00000 003 00001 003
00000 FFF 00001 FFF
00001 000 00000 000
00001 001 00000 001
00001 FFF 00000 FFF
00002 000 00003 000
FFFFF 000 00005 000To switch to process 2, the OS simply sets
cr3 to CR3_2, and now the following translations would happen:linear physical
--------- ---------
00000 002 0000A 002
00000 003 0000A 003
00000 FFF 0000A FFF
00001 000 12345 000
00001 001 12345 001
00001 FFF 12345 FFF
00004 000 00003 000
FFFFF 000 FFFFF 000Step-by-step translation for process 1 of logical address
0x00000001 to physical address 0x00001001:- split the linear address into two parts:
| page (20 bits) | offset (12 bits) | - look into Page table 1 because
cr3points to it. - The hardware knows that this entry is located at RAM address
CR3 + 0x00000 * 4 = CR3:
*0x00000because the page part of the logical address is0x00000
*4because that is the fixed size in bytes of every page table entry - since it is present, the access is valid
- by the page table, the location of page number
0x00000is at0x00001 * 4K = 0x00001000. - to find the final physical address we just need to add the offset:
00001 000 + 00000 001 --------- 00001 001because00001is the physical address of the page looked up on the table and001is the offset.The offset is always simply added the physical address of the page. - the hardware then gets the memory at that physical location and puts it in a register.
Another example: for logical address
0x00001001:- the page part is
00001, and the offset part is001 - the hardware knows that its page table entry is located at RAM address:
CR3 + 1 * 4(1because of the page part), and that is where it will look for it - it finds the page address
0x00000there - so the final address is
0x00000 * 4k + 0x001 = 0x00000001
x86 Paging Tutorial Multiple addresses translate to a single physical address by Ciro Santilli 37 Updated 2025-07-16
Ciro Santilli 37 Updated 2025-07-16The same linear address can translate to different physical addresses for different processes, depending only on the value inside
cr3.Both linear addresses
00002 000 from process 1 and 00004 000 from process 2 point to the same physical address 00003 000. This is completely allowed by the hardware, and it is up to the operating system to handle such cases.This often in normal operation because of Copy-on-write (COW), which be explained elsewhere.
Such mappings are sometime called "aliases".
FFFFF 000 points to its own physical address FFFFF 000. This kind of translation is called an "identity mapping", and can be very convenient for OS-level debugging.The exact format of table entries is fixed by the hardware.
The page table is then an array of
struct.On this simplified example, the page table entries contain only two fields:so in this example the hardware designers could have chosen the size of the page table to b
bits function
----- -----------------------------------------
20 physical address of the start of the page
1 present flag21 instead of 32 as we've used so far.All real page table entries have other fields, notably fields to set pages to read-only for Copy-on-write. This will be explained elsewhere.
It would be impractical to align things at 21 bits since memory is addressable by bytes and not bits. Therefore, even in only 21 bits are needed in this case, hardware designers would probably choose 32 to make access faster, and just reserve bits the remaining bits for later usage. The actual value on x86 is 32 bits.
Here is a screenshot from the Intel manual image "Formats of CR3 and Paging-Structure Entries with 32-Bit Paging" showing the structure of a page table in all its glory: Figure 1. "x86 page entry format".
The fields are explained in the manual just after.
The problem with a single-level paging scheme is that it would take up too much RAM: 4G / 4K = 1M entries per process.
If each entry is 4 bytes long, that would make 4M per process, which is too much even for a desktop computer:
ps -A | wc -l says that I am running 244 processes right now, so that would take around 1GB of my RAM! Pinned article: Introduction to the OurBigBook Project
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