ARM bare metal developmentLast edited on Nov 1, 2014

The project

I've always wondered how programming for an ARM cpu is. So I decided to try to make an OS, written 100% in assembly for an ARM development board. I shouldn't say OS though, every time I write an OS, I really only make: memory management, scheduler, mutex, netcard driver, serial port driver and some small application to run on the "os". It's basically just to learn about the architecture of the device.

The ARMv7 architecture offers a lot of cool features that I am not using. I just want to keep things simple for now. Once I get something working good, I will go deeper in the documentation and try some more advanced stuff.

At first, I wanted to use my beaglebone black to run my OS. But then, I found out that qemu can emulate quite a few boards and it would be easier to do. By using qemu, I get the following advantages over using a real board:

  • no need to upload code to the board, I use the image directly
  • can reboot the machine easily while working remotely (no need to physically access the board)
  • very easy to peek in memory with qemu's monitor command "pmsave"
  • can use gdb to debug with qemu
  • no need for a separate bootloader. Can boot kernel directly.

I chose to use the "realview-pb-a8" emulated board in qemu. I have never seen that board, I have no idea what it is. It uses a Cortex-A8. So I was able to get a programing guide for that SoC. I started from there.

The fact that I am using qemu makes things easier but removes a lot of fun. For example, qemu boots my kernel directly. On a real board, I would need to write a bootloader (or use u-boot). I would need to initialized SDRAM, initialize clocks and "power domains" and other board initialization. QEMU boots your kernel directly into RAM and you can run from there. So I wouldn't quite call this "bare-metal" programming. I guess I could only call this project "kernel programming for a Cortex-A8".

Getting started

The first step is to create a small test and actually run it. So I created the following program:

Note the qemu command in the Makefile. This allows me to run the test using "make run". It will emulate the ARM board which is the realview-pb-a8

Board specifications

When starting development on a new board, the first thing you need to do is to get a memory map of the device. Because the board will contain sdram, sram, memory mapped peripheral IO etc... From board to board, the physical location of those elements will change. Here is the memory map for the realview-pb-a8

Physical Memory Layout

Physical addressDescription
0x00000000-0x0FFFFFFFSDRAM mirror
0x10020000-0x1005FFFFBoard specific stuff that I don't need just yet
0x10060000-0x1007FFFFOn board SRAM
0x10080000-0x6FFFFFFFBoard specific stuff that I don't need just yet
0x90000000-0xFFFFFFFFBoard specific stuff that I don't need just yet

A more detailed memory map can be found in the RealView Platform Baseboard for Cortex A8 User Guide.


Interrupt vector table

This architecture only uses 7 interrupt vectors

0x04Undefined Instruction
0x08Software Interrupt
0x0CPrefetch Abort
0x10Data Abort

The interrupt vector table must be placed at the begining of the memory. Each entry is 32bits wide. It must be an instruction not an address. So you would typically put a branch instruction to jump to the proper handler. Using qemu, my kernel gets loaded at 0x70010000, so putting the IVT at the begining of my kernel would not work. I had to rellocate the IVT to 0x70000000 once the kernel was running. By the way, on that board the SDRAM starts at 0x70000000 but is mirrored to 0x00000000. Still, qemu starts execution at 0x70010000. but if the IVT is at 0x70000000, the CPU will still see it at 0x00000000 because of the mirror.

Setting up the stack

There are 6 CPU modes in this architecture. Each mode will shadow the register r13 (stack pointer). So they each need their own stack. To set those stacks, you must switch mode and set r13 appropriately. I don't set the User mode stack because this will be done on a per-process basis and System mode uses the same registers as user mode.

    msr     CPSR_c,#0b11010001           // stack for FIQ mode
    ldr     r13,=STACK_BASE_FIQ
    msr     CPSR_c,#0b11010010           // stack for IRQ mode
    ldr     r13,=STACK_BASE_IRQ
    msr     CPSR_c,#0b11010111           // stack for Abort mode
    ldr     r13,=STACK_BASE_ABORT
    msr     CPSR_c,#0b11011011           // stack for Undefined mode
    ldr     r13,=STACK_BASE_UNDEFINED
    msr     CPSR_c,#0b11010011           // stack for Supervisor mode. And we will stay in that mode
    ldr     r13,=STACK_BASE_SUPERVISOR

Memory Management Unit

Creating a paged memory system is not difficult. The MMU offers a 2 level page table system The level1 table has 4096 entries, each mapping 1Mb of virtual addresses. You could create "section" entries to map those 1Mb to physical memory directly. You would then get pages of 1Mb and only 1 table that takes 16k in memory. But if you want 4k pages, then those entries need to be "Coarse table" entries, meaning that each entry will reference a subtable (a level2 table). Each level 2 table contain 256 entries, mapping 4k of memory. So for a 4k paging system you would have 1 Level1 table with 4096 entries (a total of 16k in size) and 4096 level2 tables containing 256 entries each for a total of 4Mb in size.

Level1, Section
base addrNS0nGSAP2TEXAP0domainXNCB1PXN

Level1, Page Table (TODO)
page table addr0domain0NSPXN01

Level2, Small Page (TODO)
base addrnGSAP[2]TEXAP[1:0]CB1XN

Domains and permissions

The mmu has a concept of domains and access permissions. 16 access domains exist. In a page descriptor, we set the access bits and the domain associated with that page. CP15.register3 contains 2 bits for each domains 0 to 15. These bits determine how page access should be checked. Example: a page is associated to domain 12. CP15.register3 indicates that domain12 is Client. Therefore access permissions in the page will be checked. If domain12 was set to Manager, permissions would have been ignored.

Initializing the MMU

The first thing you need to do is setup the page tables like mentionned above. Obviously, you might want to do an identity mapping for the region of code that is currently running the MMU initialization code so that the mapping does not change after having initialized the MMU.

Level 1 Page Table

70100000  01 40 10 00 01 44 10 00  01 48 10 00 01 4c 10 00
70100010  01 50 10 00 01 54 10 00  01 58 10 00 01 5c 10 00
70103fe0  01 20 50 00 01 24 50 00  01 28 50 00 01 2c 50 00
70103ff0  01 30 50 00 01 34 50 00  01 38 50 00 01 3c 50 00

Level 2 Page Tables (all contiguous)

70104000  fe 0f 00 00 fe 1f 00 00  fe 2f 00 00 fe 3f 00 00
70104010  fe 4f 00 00 fe 5f 00 00  fe 6f 00 00 fe 7f 00 00
70503fd0  f2 4f ff ff f2 5f ff ff  f2 6f ff ff f2 7f ff ff
70503fe0  f2 8f ff ff f2 9f ff ff  f2 af ff ff f2 bf ff ff
70503ff0  f2 cf ff ff f2 df ff ff  f2 ef ff ff f2 ff ff ff

Then you need to configure the CP15 register.

CP15.reg2Translation Table Base RegisterLoad the base address of the Level1 table
CP15.reg3Domain Access control registerMCR p15, 0, , c3, c0, 0 where Rd contains the 32bits we want to write. We will use domain 0 for the kernel for now. So bit 1:0 will be set to 0b11 to allow access without checking permissions. all other domains will be set to 0b00 to unconditionally deny access.
CP15.reg1ControlSet bit0 high to enable the MMU. Do this as the last step

The following registers are also useful but not needed during initialization

CP15.reg5FSRRead this in you fault hander. It is the fault code
CP15.reg6FARRead this in you fault hander. It is the faulty virtual address
CP15.reg8Invalidate TLBUsed to invalidate the TLB. invalidate entire TLB: mcr p15, 0, Rd, c8, c7,0
CP15.reg10TLB LockdownUsed mark a TLB entry as persistent so it does not get overwritten by other entries. can increase performance for pages such as those containing interrupt handling code so that the translation is always cached. We could probably use a level1 table entry as a Section of 1Mb for the kernel and lock it down in the TLB.


I will not list the different reasons for getting a fault since this is all covered in the reference manual. Basically, if a fault occurs while prefetching an instruction then the Instruction Fetch fault will occur. If the fault occurs while accessing data, a Data fault will occur. The virtual address that caused the fault will be stored in FAR. A more detailed error code will be found in FSR


To test the multitasking system, I have created some small programs that I build separately and package in the image. The kernel loads the programs in their own process. Ideally, I would have like to create a flash image that contains the kernel at the very begining and then the programs would be appended at the end, kind of like a real hard disk with a bootloader and programs. But I could never get qemu to eumlate a flash file. Even with the "-pflash". The documentation for the board says that when the board is powered on, the flash is mapped at 0x00000000. This will shadow the sdram. To use the sdram,you must remap the flash to some other place. But I could never make that work. I tried creating a flash image and provide it to qemu with the "-pflash" option but that doesn't seem to work. Qemu always wants a kernel file to be provided. I don't know why I can just put my code in a flat binary that would act as flash and get the code running from 0x00000000. The kernel file gets loaded at 0x70010000 which is the sdram. So I am creating a image file containing the kernel and the programs that get loaded in sdram by qemu.

Programs run as domain 1, and in user mode. Their virtual mapping is:

Level1 table entries (1mb mapping)TypePermissionsDescription
0Section PL1 RWX, PL0 -Kernel code. Identity mapping
1-FFSection PL1 -, PL0 -Unmapped
100Page table PL1 RW, PL0 -peripherals, and SRAM. Identity mapping
101-1DFSection PL1 -, PL0 -Unmapped
1E0Page Table PL1 RW, PL0 -Peripherals. Identity mapping
1E1-1FFSection PL1 -, PL0 -Unmapped
200-2FFPage Table PL1 RWX, PL0 RWXProcess code
300-6FFSection PL1 -, PL0 -Unmapped
700-8FFSection PL1 RWX, PL0 -Kernel code. Identity mapping
900-EFFSection PL1 -, PL0 -Unmapped
F00-FFFPage tables PL1 RW, PL0 RWProcess Stack

The task information page

When creating a process, I add it in a list of process. The list of process is a fixed-size list in kernel memory (accessible by any process in privileged mode) that contains a pointer to the L1 table of the process and several other usefull information for the process. This information is used by the scheduler and is formatted like this:

0x0000Physical address of the process's L1 page table
0x0004saved r13_irq registers
0x0040Quantum count

Using Software Interrupts

When using the SWI instruction, you need to pass it a parameter that would normally be the function number you would want to call. For example: SWI 0x02. Once you are in the SWI handler, you want to get that parameter to know where to dispatch the handler. but the SWI instruction completely ignores the parameter. It is not given to you in any way when you get in your handler. In order to get this, you need to take r14, which contains the return address and substract 4. That would give you the address of the SWI instruction itself. So you can read at that memory area and see what parameter was provided. That is pretty weird in my opinion. I would rather just put the function number in r0 before calling SWI and read r0 once in the handler. That would illimnate an unncessary memory access. Plus, the page at that location will obviously be in the instruction prefetch cache but since we are using "LDM" to load the instruction in a register to read it, it means we will be looking in the data cache. And the page will most probably not be in that cache. So in my project, I will only pass function parameters in a register and ignore the one provided to SWI.

Something that confuses me is that when calling SWI, you enter Supervisor mode. Then you are in privileged mode. That makes sense, But then r13 and r14 gets shadowed. I'm not entirely sure why I would want that. It actually complicates things when multi-tasking. I guess that in some more complex OS design, this is very usefull.


Saving registers of mode X from mode Y

Assume we have a function called schedule(). This function saves the current context, and reloads the context of the next task to run. In my implementation, this function willa lways be called from the IRQ mode. So the schedule function will be called from a non-user mode. The schedule function will need to store the user-mode context (registers r0 to r14). But from the non-user mode of IRQ, registers r13 and r14 are shadowed. r0-r12 will be the same as the user-mode so we need to find a way to save the r13 and r14 of the user mode. For this, the instruction stm/ldm with "^" can be used to store/load the user-mode registers. this will save/load r0-r12 as usual but the r13 and r14 will be the ones of the user mode.

CPSR and SPSR: While in an exception (therefore in a mode different than user or system) the previous cpsr is saved in spsr. Before returning back from the exception, you must reload spsr back into cpsr. This will change the mode automatically, re-enable interrupts etc. To load this and to load r14 in r15 at the same time, look at the notes below about the LDM instruction.

LDM instruction format: Compared to AVR32 and x86, this is pretty complicated in my opinion. The "ldm" instruction has 3 forms. The first form does what it says it does. But the second form which is: ldm Rn,registers_without_r15^ (yes, there is a "^" at the end) loads all user mode registers while you are in a non-user mode. so it is a way to load user registers while they are shadowed. The third form, ldm Rn,registers_with_r15^, will automatically load spsr back into cpsr. You could also use a data instruction with the "S" flag and R15 as a destination. For some reason, it will conveniently reload spsr back into cpsr at the same time... Go figure. For example: movs r15,14; will reload r15 and also reload spsr back into cpsr. I am wondering why they re-purposed a flag like this. That is one small thing that makes me like x86 more than ARM.

Since I re-enable interrupts after entering SWI, the SVC context must be saved also since a context switch could occur while in a service call (that is actually the whole point of re-enabling interrupts in SWI). So my context-switching code also pushes the r13_svc and r14_svc on the the task's IRQ stack.

Context switching

The schedule function needs to do the following:

  • save registers r0-r14 user-mode
  • save register r14_irq (since this will be done from the IRQ handler)
  • save register spsr (which is the usermode cpsr)
  • change level1 page table for new process
  • flush tlb (unless using ASID)
  • restore r0-r14 (for user-mode)
  • restore the return address in r14_irq
  • restore spsr
  • to return, load r14_irq in r15 and spsr into cpsr

Each task have their own stack and they have their own IRQ mode stack. When entering the schedule() function in IRQ mode, the use-mode registers are pushed onto the IRQ stack. The current spsr and r14_irq is also pushed on the IRQ stack. The r13_irq is then saved in a list somewhere. When time comes to restore the task, the page tables are switched back to that task's page tables, and r13_irq is restored from the list. At this point, the task's IRQ stack has been restored. We can then pop everything from it and the context switch is done. Here is a sample of my schedule function

    mrs     r0,SPSR
    // At this point, whole context is saved on stack

    // Determine what is the next task to run

    // store context: save r13_irq
    // r0 points to the entry in the process list (as decribed earlier)
    str     r13,[r0,#4]     // offset 0x04 is r13

    // load new page table
    ldr     r1,[r5]             //r5 points to the entry of the next task to run
    mcr     p15,0,r1,c2,c0,0
    // flush TLB (note that there are ways to avoid this
    mov     r1,#0
    mcr     p15,0,r1,c8,c7,0

    //load r13_irq
    ldr     r13,[r5,#4]

    // now restore context on stack
    pop(r0)            // this is just SPSR, only reg available to touch is r14
    msr     SPSR,r0
    b       returnFromInterrupt

This is a sample only. My schedule() function does a bit more than that. But it gives you the general idea.

When r0-r14 will be restored for the user mode, it would restore the task's context as it was before entering the IRQ to schedule(). r13 and r14 of the user mode will be restored and not the banked ones of the currently executing mode.

Note that the TLB must be flushed when reloading the "translation table base register" in CP15 because the cached TLB entries will continue to correspong to the previous mapping. This is a very expensive operation but we can use the concept of ASID by using the CONTEXTIDR register. By setting a unique task ID in CONTEXTIDR, all page translations that gets loaded in the TLB will be tagged with that ID. When doing a lookup, the MMU will ignore entries that do not match the current CONTEXTIDR. So on a context switch, you would change the ID in CONTEXTIDR. This would create duplicate entries in the TLB but with different IDs. So instead of flushing the TLB, entries will be removed only when the TLB is full. See this article for more information about the TLB and ASID.

The schedule() function is called by the timer IRQ handler. But you might want to call it from other places. For example, if a task wants to yield, it should be scheduled out immediately. I could do this in a SWI handler but trying to change context from the SVC mode brings up other challenges. So to keep things simple, I want to do context switches only from the IRQ mode. For this, it is possible to use a "software IRQ". This is well documented in the GIC documentation.


Here is my source code

Using epollLast edited on Oct 10, 2014


I heard about epoll not so long ago and wanted to give it a try. Epoll is the new way to use non-blocking sockets (or any file descriptors). It replaces poll() and select().

epoll is easy enough to use. The good thing about it is that, unline select(), you don't need to rebuild the FDSET each time you call epoll_wait. Here is a typical usage:

  • efd = epoll_create1()
  • epoll_ctl(efd, EPOLL_CTL_ADD, listeningSocket)
  • loop:
    • n = epoll_wait()
    • for i=0 to n:
      • if (events[i].data.fd == listeningSocket)
        • newsock = accept()
        • epoll_ctl(efd, EPOLL_CTL_ADD, newsock)
  • close(efd)

Another nice thing about epoll is that you don't need to worry about removing a socket from the list. You can call epoll_ctl(efd, EPOLL_CTL_DEL, sock) if you want but when the socket closes, it will be removed automatically.

One thread

Using epoll, I can do all my sending and receiving in one thread. So people may suggest to send data from other thread but what if you get EAGAIN? Assume that thread A is the epoll thread. Thread B attempts to send 100 bytes but could only send 50. Then Thread C attempts to write 100 bytes on the same socket. By the time that Thread C was called, the socket was ready. The remote socket will have received corrupted data. For that reason, Thread B and C, will add data in a queue so that each messages are guaranteed to be sent completely. The queue will be emptied in Thread A.

Since epoll might potentially handle thousands of connections, Thread A must do minimal work. It must only do this:

  • epoll_wait
  • if a socket is ready-ready, read all from that socket and dispatch to consumers. consumers should work fast, otherwise we should "post" the data to the consumer in another thread.
  • send all from the sendqueue

Edge-triggered VS Level-Triggered

epoll offers two ways of working: Edged-triggered and Level-triggered. The default way is Level-triggered. so you might want to change this to level-triggered in your application.

In Level-triggered mode, epoll will return as long a socket is read or write ready. So if data is ready to be received on the socket and you only read part of it, the next epoll call will return because there is still data available. As soon as the internal receive buffer is empty, then epoll won't return. But since the socket will be write-ready almost all of the time, it will return. This is not something you would typically want. My guess is that if you want to use level-triggered mode, you should not register to get EPOLLOUT events unless you have something to send. So while your application's TX buffer is empty, you don't register for EPOLLOUT. As soon as data is added to the TX buffer, you register for EPOLLOUT, epoll will return and you write data out on the socket. If EAGAIN was returned then you will block on the next epoll_wait() and can send the rest of the buffer when it unblocks. Once the application's TX buffer is empty, you could unregister for EPOLLOUT.

With Edge-triggered mode, things are different. You will only receive an event if the status has changed from not-ready to ready. So if you have incomming data and you only read part of it, when you will call epoll_wait, the socket will still be ready. So epoll_wait will block because the state will not change from not-ready to ready. So if you are using that mode, you must make sure to read ALL the data on the socket until you get EAGAIN. If you want to send something, epoll_wait will only give you a write-ready event if you previously got EAGAIN while attempting to write out.

Let's say you have an internal transmit queue in which you add data when you want to send it out. The epoll thread would need to ready from that queue, write the data on the socket and handle EAGAIN appropriately.

  • epoll_wait()
  • if a socket is ready-ready, readAllFromThatSocket(socket);
  • sendAllData() // sends all that is contained in the internal queue.
  • goto step 1

But if epoll_wait() is blocked in triggered mode, and you add data in the queue (from another thread), sendAllData() will not be called until epoll_wait returns because data is ready to be received (it won't return because data is ready to write, because you need to write first and get EAGAIN for that.). To solve this problem, I created an eventfd (see sys/eventfd.h). I add the eventfd in the epoll_wait list and whenever I add data in the application's TX queue, I write 1 on the eventfd so that epoll_wait will break out. I could have used a timout in epoll_wait, but that would add unnecessary latency to amy reactor. That way of doing things is similar to what is called the "pipe to self trick".

My framework

To play with epoll, I wrote a small reactor framework that exposes an interface that's easy to use. all the gory details are hidden inside the framework. To use it you would need to define two classes:

class OwnProtocolFramingStrategy: public IFramingStrategy
    virtual int processData(char* buffer, size_t size, FullMessageInfo& info)
        info.buffer will first be zero. You need to allocate memory for that buffer
        and copy the data from "buffer" to "info.buffer". If you determine that the buffer
        does not contain a full message, return the quantity of bytes you read from the buffer. 
        In this case, it should be "return size;" And the next time that some data is received,
        this function will be called with you buffer. You can update other fields in "info"
        to help you resume next time.

        If you determine that a full message was received, set "info.complete = true" and return
        the number of bytes that you read from the buffer. After returning from this function, your
        buffer created in "info.buffer" will be considered as containing a full message and will be 
        passed to the TcpEngine. Next time this function will be called, info.buffer will be back to 

        If you determine that a full message was received but there are still data left in the buffer,
        it means that you probably have 2 messages in the buffer. As mentioned above, you will 
        return the number of bytes that you read from the buffer. This function will be called
        again with "buffer" pointing to the index after the last byte you have read. So you
        only need to try to parse one message only when this function is called.

        If you find that data is not well-formed, does not respect you protocol or has any other errors 
        that prevents you from reliably build a full message, then return -1 and the client's connection
        will be aborted.        

A IFramingStrategy is a class that will process the tcp stream into full messages. As you know, when reading data from a socket, you may receive more than one message in a single read and you could also only receive half of a message. So the IFramingStrategy is what parses the stream and builds messages. For example, if implementing an HTTP Server, the strategy would build the message by expecting the data to be chunked so the logic to calculate the size would be done here.

class OwnProtocolServerEngine: public TcpEngine
    OwnProtocolServerEngine():TcpEngine(new ClientFactory(this))

    virtual void onClientConnected(TcpClient* client)
        Maybe add the client in some list so we can use it later?

    virtual void onClientDisconnected(TcpClient* client)
        if client was added in a list, we should remove it because
        its instanced will be deleted after returning from this method.

    virtual void onClientMessage(TcpClient* client, char* buffer, size_t size)
        The buffer you get here, is the one that was created in the strategy. You
        own this buffer in this method, so you must free it from here 
        (or remember to free at a later time)

        In this method, you are guaranteed to get 1 and only 1 full message assuming
        that the Strategy was coded correctly.

        Note that everything is happening in 1 thread. So in the onClientConnected() 
        method, you might have saved a TcpClient pointer in a list. It is perfectly
        safe to use it in here (i.e: for forwarding the message to someone else).

Then to run launch a server:

OwnProtocolServerEngine engine;
if (!engine.start())
    printf("Could not start server\r\n");
    return -1;


The source code for the framework is here

Thread-safe queue

I said earlier that everything was happening in 1 single thread so everything *should* be safe. But what if I want to send data to a client from another thread? That should be possible. The transmit queue inside the TcpClient is thread safe. So calling TcpClient::sendData() is a safe call to do in another thread. Accessing the client is another story though. The client instance could get deleted at any time, but there are ways around that but it's beyond the scope of this framework.

Since the TX queue is a Multi-Producer, Single-Consumer FIFO, creating a thread-safe implementation is very easy. Of course, I didn't want to use any of the STL containers.

This is using __sync_bool_compare_and_swap, which I strongly suspect uses the x86 instruction CMPXCHG. I described that instruction here I didn't fully test that code though . I addapted it from a version I had written in ASM, so there might be bugs in the C++ version.

Memory PagingLast edited on Sep 26, 2014


I would like to describe how memory paging works. So far, I have played with x86 paging, x86_64, ARM, and avr32 paging. All 4 architectures are different but they share some fundamental concepts that I will try to describe here. There are many ways to implement a paging mechanism, but I am only sticking to the basics here. In my opinion, this article is a good place to start if you want to understand the basics of paging before moving to the technical details and advanced algorithms.

Mathematics of pages

When an instruction attempts to access a memory location, let`s say at address 0x00002345, the MMU will treat that as a logical address. It will separate the address into a page number and an offset. If using a paging system with a 4k page granularity, pages will be 4k each. By taking the address and dividing it by 4096, you get a quotient of 0x02 and a remainder of 0x345, so the mmu would determine that the address points to page #2 at offset 0x345 within the page. The fact that page sizes are always powers of 2 is very convenient because instead of doing a division by 4096, you can simply extract the right-most 12 bits as the offset and all upper bits as the page number. So page_number = address>>12. offset = address&0xFFF.

Note that when using a logical address, you don't need to think about the page number and offset. Using address 0x00002345 will yield page 0x2, and offset 0x345, and if you go 4k above that, the address would be 0x00003345, so page 0x3 and offset 0x345. This is all done naturally.

Page tables

Page tables are tables of descriptors residing in memory. Each descriptor is, for example, 8bytes long, or 64bit. Each descriptor are in order of pages. Descriptor 0 contains information about page 0 (0..4095), Descriptor 1: page 1 (4096..8191), Descriptor N: Page N (N*4096..(N+1)*4096-1). A descriptor is a mapping to physical memory, it contains a physical address. So let's say you need to access logical address 0x0000000000003DEF, the MMU will determine that you want to access page 3, offset 0xDEF. The 3rd descriptor could hold the value 0x0123456789ABC000. So that means that the MMU will map the logical address 0x0000000000003DEF tp physical address 0x0123456789ABC000 + offset 0xDEF, so 0x0123456789ABCDEF. See we have just transposed the offset over the physical address? That is no coincidence. It is because the addresses in the descriptors will always be a multiple of 4096, since pages can only start on a 4k boundary (for 4k pages that is). Therefore, the 12 lower bits of the descriptor will always be zero. For that reason, those bits are always reused to contain other information about the page such as permissions etc... But that is out of the scope of this article. Basically, what this all means is that the mapping is like this: PhysicalAddress = (PageTable[LogicalAddress>>12] & ~0xFFF) | (LogicalAddress&0xFFF).

The mapping is done transparently by the CPU!

So far, paging is cool, but here is why it is powerfull: There can be many page tables in memory and at any moment, the MMU can be instructed to use another table to change the mapping. Usually, in an operating system, you want to use 1 page table per process. Let's say you write an operating system, and to make things simple, you want the heap space of each process to begin at 0xBEEF1000. That is easy enough with paging because you just create a page table for each of the processes and out an address in page number 0xBEEF1. Obviously, you don't want each process to overwrite the data of the other processes, so the page 0xBEEF1 in each different page tables would point to a different physical location.

Note that if you only have 1g of RAM, the last accessible memory location would be 0x0FFFFFFF. But there is no limit to the logical addresses (other than the architecture size, 32bits or 64bit) as long as they map to a valid physical location (that is, under 0x40000000 if you have 1gig).

Since each process can have a full page table to themselves, it means that, as far as they know, they have a full 4gig of RAM (in the case of 32bit architecture) that they can access. so let's say that you have 4 process and 4 gig of RAM.

  • Process 1 maps the first gig of logical addresses to the 1st gig of physical addreses.
  • Process 2 maps the first gig of logical addresses to the 2nd gig of physical addreses.
  • Process 3 maps the first gig of logical addresses to the 3rd gig of physical addreses.
  • Process 4 maps the first gig of logical addresses to the 4th gig of physical addreses.

If process 1 needs more memory, it sees that only the 1st gig of memory is used (because it only deals with logical addresses and not physical addresses). So it will naively think that there is 3 gig of RAM left. It will ask the kerl to map another gig of ram, but the kernel will say: Wait! there is no more physical memory to which I can map these logical addresses. Well that's not entirely true. This is where swapping comes in.

Page Swapping and Page faults

When the kernel runs out of physical memory so map to logical memory for a process, it will re-use a physical location. There are many ways of determining which physical location to use, but for simplicity, let's assume that this would be done randomly. As per our example above, all 4 process uses all the memory. Process1 requests the kernel to map some physical memory to logical address 0x40000000. The kernel has no more physical memory to hand out so it will pick a random location and will choose physical address 0x80000000 for example. This address is mapped to Page 0 of Process 3 already. So the kernel will modify the page descriptor 0 in page table of process 3 and mark it as invalid. Remember the low 12bits that are unused in the descriptor and I said there would be used for information about the descriptor? Well the "invalid" bit is one of those 12 bits. The kernel will then take the full 4k residing at that physical location and store it on the hard drive, in a swap partition (or file). It will then map the newly requested logical space for process 0 to that physical location. At this point, we have 2 process that have 2 different pages that points to the same pysical location. Only one of those process has the page marked as being valid in its descriptor.

When execution resumes on process2, it could attempt to access page 0 of its logical space. When it does, the MMU will see that the page is marked as invalid and will trigger a Page Fault exception. The kernel will then handle the page fault exception. It will swap the 4k in memory with the one that was previously stored on the hard drive. mark the page in fault as valid and mark the other one as invalid. And that is how swapping works.


Another use of paging is for memory fragmentation. Let's say that you have 2 process. and only 16K of memory, so 4 pages of 4k. Process 1 is uing 2 4k buffers and has a mapping like this:

  • Page0 = 0x0000
  • Page1 = 0x2000
  • Page2 = unused
  • Page3 = unused

it means that Process 1 is using 8k of RAM. There is only 8K of RAM left. Process2 comes in and wants to create a buffer of 8K. Obviously, it expects that buffer to be contiguous memory. But that would be impossbile since there are only 2 4k chunks left and they are not contiguous. Pagin will solve that. Process2 will have a mappint like this:

  • Page0 = 0x1000
  • Page1 = 0x3000
  • Page2 = unused
  • Page3 = unused

When process2 will access logical address 0x0000..0x0FFF, it will be transparently redirected in 0x1000..0x1FFF and then when it continues to 0x1000...0x1FFF, it will be redirected in 0x3000..0x3FFF. So the logical memory IS contiguous. This concept makes it very easy to avoid memory fragmentation as we allocate/deallocate memory dynamically.

Translation Lookaside Buffer (TLB)

When an access to memory is requested, the CPU needs to translate the virtual address to physical address. Like I mentionned above, this is done by walking through the page tables. This process adds a significant ammount of time to the memory access because the CPU need to make one or several memory access just to decode the address. So we are talking about more than 100% overhead here. To reduce the impact of this process, most CPU (if not all) implement what is called the Translation Lookaside Buffer, or TLB. When a virtual address is translated, the result is stored in the TLB. So when the next memory access is requested, the CPU will start by looking in the TLB. If a match is found, then it doesn't need to look into the page tables. The TLB is inside the CPU, so accesses to it is very fast. But the TLB has a limited size. When a new virtual address is translated, and the result cannot be stored in the TLB, the CPU will choose a "victim" entry (the word victim is actually a word that some official documentation use) and delete that entry to replace it with the new one. The algorithm for choosing the victim depends on the CPU implementation, but would mostly be a "least recently used" type of algorithm. The TLB works kind of like a DNS cache but without TTL.

When a context switch occurs, the new running process might want to use a different page table for its own translation. That is actually one of the many benefits of using a virtual memory system: so that each process can have their own translation tables. The OS would then instruct the CPU to use another table by giving it the base address of the new table. But the TLB still holds some translation that might not match the new table. We call those entries "stale entries". So when updating the Page Table Base Address, we must also flush the TLB to remove all stale entries. But flushing the TLB could add a significant ammount of processing. And what if the new process only needs 1 translation, and then we context-switch back the the old process. There could be enough place in the TLB to store all the needed translation for both process. So flushing the TLB is very inneficient in that case. That is why most CPUs will implement the concept of an Address Space ID (ASID). On a context switch, ou would update a special register in the CPU to tell it the ID of the currently running process (that would be the ASID). When the CPU adds an entry in the TLB, it will add a "tag" to the entry, that tag would be the ASID. When a TLB lookup is performed by the CPU, any entries tagged with an ASID that does not match the current ASID will be disregarded. So it would be possible to have two identical translations but with different ASID. This illiminate the need to flush the TLB on every context switch because stale entries will be tagged with a different ASID anyway. When the TLB becomes full, the victims will preferably be entries for other ASID than the current one.

Some translations might need to be performed very often. Let's take, for example, the page containing the code for interrupt handlers. That page needs to be accessed very fast, and we would want it to stay in the TLB all the time. Most CPUs offer the possibility to mark some TLB entries as "permanent" or "locked". These entries will never be chosen as victims if the TLB becomes full.

I've seen architectures, such as the AVR32, that doesn't do automatic TLB bookeeping. When a TLB miss occurs, they triger an exception and it is the duty of the OS to fetch and decode the page descriptor and load it in the TLB.

Addressing Space Limitation

Some people may wonder, if my CPU is a 32bit CPU, why can't I use 4gig of RAM?. This is because a 32bit CPU can only use a 32bit integer as the address of a memory location. So it could address 4 gig of RAM theoritically. But you might have a video card in your computer that has 1gig of RAM on it. That 1gig must be accessible by the CPU. So if you have 4 gig of RAM and 1gig in the video card, it means that your CPU would need to access a total of 5gig of ram, but that is impossible. You video card's RAM will be assigned a physical range of addresse. The 32bit adressing really just defines an addressing space. This is kinda like if you were living on a very long street but there is a rule in your neighbourhood that prevents you from using more than 2 digits on on the address plate of your house. Even though about 1000 houses could be built on that street, after the 100th house, the next ones couldn't be assigned a house number because the addressing space was all used up. In my example, I talk about a video card, but there are many other peripherals that uses addresses from the addressing space.

Then you might wonder: how come I have a 32bit CPU and I can access my full 4gig on linux or windows 7? This is because newer (they are actually old at the moment) CPUs have a 36bit addressing mode even though the CPU is 32bit. But the OS must make use of that. WindowsXP did not use the 36bit addressing mode even if the CPU offered it. That is why it was impossible to use your whole RAM on windowsXP

Implementing HTTP Digest AuthenticationLast edited on Aug 15, 2014

Recently, I was trying to add HTTP digest authentication on my Home automation device. The device exposes a REST interface trough a proxy server. My web server is setup like this

Now since my API is exposed to the world by proxying it like that, I wanted to add security by implementing HTTP digest authentication. Whether or not Digest authentication with MD5 is secure or not is a completely different story, but let's assume it is good enough for now. I have a restricted access webpage that I go on to control my home automation device. This web page makes requests to DHAS using javascript. Since I've implemented digest authentication, I now need to put the credentials in the javascript so that the calls made with XMLHttpRequest can succeed. Even though that javascript code will only be served to me, while I am authenticated on the website, I felt uncomfortable to leave a hardcoded username and password in the JS source. So this is what I came up with:

Note that messages sent from JS to DHAS are being proxied by Apache. Therefore, DHAS receives a GET for /insteon/listmodules and not for /dhas/insteon/listmodules

  • use XMLHttpRequest to make a request to DHAS (through the proxy)
  • add a header "X-NeedAuthenticationHack" in the request
  • receive a 401
  • get the "X-WWW-Authenticate" header from the 401 response
  • Make a XMLHttpRequest to the server and send it the "X-WWW-Authenticate" data
  • Server side php script with hardcoded username/password for DHAS solves the challenge and returns the resonse
  • use XMLHttpRequest to make a request to DHAS (through the proxy) and append the response in a "Authorization" header

So basically, I just intercepted the 401 and instead of letting the browser prompt for a username password, I created the response myself. And instead of doing in the JS, I did it on the server, limiting the exposition of the username/password. You may notice my two special X headers. This is because if the server returns a 401 with WWW-Authenticate, the browser will prompt for your credentials. Event if I have a handler defined to get the 401. So when I send my initial request, I set the X-NeedAuthenticationHack header to tell the server: "Hey, don't send me a WWW-Authenticate, send a X-WWW-Authenticate instead so I can deal with it".

By the way, even if the information is easy to find, this is how the digest authentication is done:

  • Client makes request to http://webserver.com/url1/index.html
  • Server sends a "WWW-Authenticate: realm="testrealm", nonce="testnonce"
  • ha1 = md5("username:testrealm:password")
  • ha2 = md5("GET:/url1/index.html")
  • ha3 = md5(ha1+":testnonce:"+ha2)
  • Client sends: "Authorization: Digest username="username", realm="testrealm", nonce="testnonce", response=""+ha3+"", uri="/url1/index.html"

Stack frame and the red zone (x86_64)Last edited on Mar 18, 2014

after days, and days, and days of troubleshooting odd problems I had in my homebrew x86_64 OS, I found out that it was caused by my C compiler. After spending all these years arguing with everyone that it is easier to make a hobby OS in pure assembly, I decided to make this OS with asm and C, and I just proved myself that C is evil! Seriously, C is not evil but it does hide a lot of things that makes it hard to know what your OS does.

So after disassembling all my C code and inspecting the assembly code, I found this:

55                push rbp
4889E5            mov rbp,rsp
48897DE8          mov [rbp-0x18],rdi
488975E0          mov [rbp-0x20],rsi
488955D8          mov [rbp-0x28],rdx
C745FC00000000    mov dword [rbp-0x4],0x0
C9                leave
C3                ret

Notice how the function never decreases the stack pointer? Arguments are passed below the stack pointer. Can you image how insane that is???? What would happen if an interrupt would trigger while in that function? The correct code would be:

55                push rbp
4889E5            mov rbp,rsp
4883EC28     ---> sub rsp,byte +0x28 <---
48897DE8          mov [rbp-0x18],rdi
488975E0          mov [rbp-0x20],rsi
488955D8          mov [rbp-0x28],rdx
C745FC00000000    mov dword [rbp-0x4],0x0
C9                leave
C3                ret

It turns out that this behavior is normal according to the amd64 ABI. There is a thing called the "red zone". The red zone is a 128 bytes buffer that is guaranteed to be untouched by interrupt handlers (I'm not sure how though). To quote the ABI:

The 128-byte area beyond the location pointed to by %rsp is considered to be reserved and shall not be modified by signal or interrupt handlers. Therefore, functions may use this area for temporary data that is not needed across function calls. In particular, leaf functions may use this area for their entire stack frame, rather than adjusting the stack pointer in the prologue and epilogue. This area is known as the red zone.

So my solution was to just disable that damn red-zone with the gcc flag "-mno-red-zone". I'm guessing that the compiler does that to improve performances because it assumes that your code will be running in ring-3, so when an interrupt occurs, the stack will change because the handler will run in ring-0. Yeah sure, it will improve performances because there is one less instruction in the code, but I think that's a huge assumption to make. It definitely isn't the case when you are writing kernel code anyway.