We have learned that the information in the hard disk are never erased, only the links to it are erased. Does this apply to RAM as well (as long as the computer is on of course)?
For example, When I declare a variable before I give it a value, the value can be 0 or some other value that was previously in that place.
Can I say that the computer doesn't know how to delete something it just replaces it with another?
When a process (program/app) starts, it asks the operating system for some memory. The operating system keeps track of the address/size of the memory each process uses. A process can't access a region of memory it doesn't own (virtual address mapping). When the process ends the operating system may choose to write a value (like 0x00) to all the memory a process had. I believe most existing operating systems don't overwrite memory.
Take a look at this question for a further insight about this subject.
Related
Context:
I don't really understand how the kernel saves the state of a running code when it gets to exceed its time slice.
I don't visualize what happens actually.
Question:
1) Where is stored the current running code (and its stack ?) ?
2) When the kernel will "see" the code again, will it just follow an offset and keep going as if nothing happened ?
It is not clear to me.
Thanks
Current code instruction pointer and current stack pointer are stored in task_struct->ip and task_struct->sp (for x86) and new process's task_struct->ip and task_struct->sp and are loaded back to sp and ip registers when switch_to() is called in Linux kernel.
Kernel's switch_to() does many things like resetup of EIP, stack, FPU, segment descriptors, debug registers while switching to new process.
Then kernel's switch_mm() switch the virtual memory mappings from last process to new process.
It depends on the OS but as a general rule there is a block of storage which holds information about each process (usually called the Process Control Block or PCB). This information includes a pointer to the current line of code that is being executed and the contents of registers etc, so the process can start again where it stopped last time.
This block of information is owned by the OS itself not the process so it lives beyond the suspension of the process.
The program code itself is not stored in the PCB - it simply exists in memory or on disk. It can even be shared between processes, for example several processes may be running the same program, each at a different point in the code at any given time and each with their own set of 'variables' or data unique to that process's run of the program. All the OS needs is the variables and the line number or pointer to know where a particular process was in the code when it was suspended, and it can start from that point again.
It is worth noting that any RAM the process was using may or may not be still there when it restarts. In general an OS will try to leave recently used or frequently used RAM chunks (or 'pages') in memory if possible. If it needs to free up space, however, it may swap the 'page' out to disk, but disk access is much, much slower, hence the desire to avoid swapping out memory which is likely to be used again if possible.
In the worst case situation an OS may find it swaps out a process and then very soon the new process need to use some memory which has to be retrieved from disk. It is suspended while this happens as the retrieval take a long time in CPU terms. It may then happen that the next process also very soon finds itself in the same situation. The OS is now spending a lot of its time swapping processes and memory in and out and much less of its time doing real work - this is commonly called 'thrashing'.
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Per my understanding, virtual memory is as follows:
Programs/applications/executables reside in a storage device. Storage device access is much slower than RAM. Hence, programs is copied from storage memory to main memory for execution. Since computers have limited main memory (RAM), when all of the RAM is being used (e.g., if there are many programs open simultaneously or if one very large program is in use), a computer with virtual memory enabled will swap data to the HDD and back to memory as needed, thus, in effect, increasing the total system memory.
As far as I know, most embedded devices do not have disk memory (like smartphones or in car infotainment systems). Code is directly executed from Flash memory. RAM is mainly used as a scratchpad area (local variables, return address etc).
So why do we need virtual memory in embedded systems? (e.g. WinCE and QNX support virtual memory)
Your understanding is completely wrong. You are confusing virtual memory with swapping or page files. There are systems that have virtual memory and no swap or page files and there are systems that swap without virtual memory.
Virtual memory just means that a process has a view of memory that is different from the physical mapping. Among other things, it allows processes to have their own virtual address space.
Storage device access is much slower than RAM. Hence programs is copied from storage memory to main memory for execution. Since computers have limited main memory (RAM), when all of the RAM is being used (e.g., if there are many programs open simultaneously or if one very large program is in use), a computer with virtual memory enabled will swap data to the HDD and back to memory as needed, thus, in effect, increasing the total system memory.
That's swapping (or paging). It has nothing to do with virtual memory except that most modern operating systems implement swapping using virtual memory. Swapping actually existed before virtual memory.
I think you're probably incorrect about these devices running code directly from flash memory. The read speed of flash is pretty low and RAM is very cheap. My bet is that most of the systems you mention don't run code directly from flash and instead use virtual memory to fault code into RAM as needed.
embedded systems, the term itself has a wide range of applications. you could call a small microcontroller with flash program space measured in kbytes or less and ram measured in either bits or bytes (not enough to be kbytes) an embedded system. Likewise a tivo running a full blown operating system on a pretty much full blown computer motherboard (replace tivo with xbox as another example) as an embedded system. So you need to be less vague about your question. virtual memory has little to do with any of that its applications cross those boundaries.
There are many answers above, David S has the best of course that virtual memory simply means the memory address on one side of the virtual memory boundary is different than the physical address that is used on the other side of that boundary. Where, how, why, etc is there a boundary varies.
A popular use for virtual memory, and I might argue a primary use case is for operating systems. One benefit is that for example all applications could be compiled for the same address space, all applications might be compiled such that from the programs perspective they all start at say address 0x8000, and as far as that program when it runs and accesses memory it accesses stuff based on that address. A combination of the hardware and the operating system change that virtual address that the program is using to a physical address. If the operating system allows for multitasking, then each task might think they are in the same address space but the physical addresses are different for each of those tasks. I wont elaborate further on why using an assumed, fixed address space, is a benefit. Another aspect that operating systems use is memory management. Many MMU's will let you segment the memory however. If a user wants to allocate 100 Megabytes of memory the program may access in its virtual address space that 100 meg as if it were linear and in that address space it is linear, but that 100 meg might be broken down into say 4Kbyte chunks that are scattered all about the physical address space, not always likely but certainly technically possible that no two chunks of that physical memory is next to any other chunk of that 100 meg. your memory management doesnt necessarily have to try to keep large physical chunks of memory available for applications to allocate. Note not all MMUs are exactly the same and 4Kbytes is just an example. A third major benefit from virtual address space to an operating system is protection. If the application is bound to the virtual address space, it is often quite easy to prevent that application from touching the memory of any other application or the operating system. the application in this case would operate/execute at a proection level such that all accesses are considered virtual and have to go through a translation to physical, the tables that are used to define that virtual to physical can contain protection flags. If the application addresses a memory address in its virtual space that it has no business accessing, the hardware can trap that and let the operating system take action as to how to handle it (virtualize some hardware, pop up an error and kill the app, pop up a warning and not kill the app but at the same time feed the app bogus data for their transaction, etc).
There are lots of ways this can be used in an embedded system. first off many embedded systems run operating systems, so all of the above, ease of compiling the program for the address space, relative ease of memory management, and protection of the other applications and operating system and other benefits not mentioned. (virtualization being one, being able to enable/disable instruction/data caching on a block by block basis is another)
The bottom line though is what David S pointed out. virtual memory simply means the virtual address is not necessarily equal to the physical address, it can be but doesnt have to be, there is some boundary, some hardware, usually table driven, that translates the virtual address into a physical address. Lots of reasons why you would want to do this, since some embedded systems are indistinguishable from non-embedded systems any reason that applies to a non-embedded system can apply to an embedded system.
As much as folks may want you to believe that a system has a flat address space, it is often an illusion. In a microcontroller for example you might have multiple flash banks and one or more ram banks. Each of these banks has a physical, generally zero based address. Even if there is no mmu or anything else like that there is a place somewhere between the address bus on the processor and the address bus on the flash or ram memory that decodes the address on the processor and uses that to address into the specific memory bank. Often the lower bits match and upper bits are responsible for the bank choices (this is often the case with an mmu as well) so in that sense the processor is living in a virtual address space. (not limited to microcontrollers, this is generally how processors address busses are treated) With microcontrollers depending on a pin being pulled high or low or some other mechanism you might have a chip feature that allows one flash bank to be used to boot the processor or another. You might tie an input pin high and the processors built in bootloader allows you to access and debug the system for example reprogram the application flash. Or perhaps tie that line low and boot the application flash instead of the vendors debugger/boot flash. some chips get even more complicated letting you boot one flash then the program writes a register somewhere instantly changing the memory architecture moving things around, for example allowing ram to be used for the interrupt vector table so your application can be changed after boot rather than a vector table in flash that is not as easy to change at will.
now when you talk about virtual memory as far as swapping to and from a disk, that is a trick often employed by operating systems to give the illusion of having more ram. I mentioned that above under the category of virtualization. virtual memory in the sense that it isnt really there, I have X bytes but will let the software think there are Y bytes (where Y is larger than X) available. The operating system through the virtual tables used by the hardware, manages which memory chunks are tied to physical ram and are allowed to complete as is by the hardware, or are marked as not available in some way, causing an exception to the operating system, upon inspection the operating system determines that this is a valid address for this application, but the data behind this address has been swapped to disk. The operating system then finds through some algorithm another chunk of ram belonging to whomever (part of the algorithm) and it copies that chunk of ram to disk, marks the table related to that virtual to physical as not valid, then copies the desired chunk from disk to ram, marks that chunk as valid and lets the hardware complete the memory cycle.
Not any different than say how vmware or other virtual machines work. You can execute instructions natively on the hardware using virtual memory until such time as you cause an exception, the virtual machine might think you have an xyz network interface and might have a driver that is accessing a register in that xyz network interface, but the reality is you have no xyz hardware and/or you dont want the virtual machine applications to access that hardware, so you virtualize it, you trap that register access, and using software that simulates the hardware you fake that access and let the program on the virtual machine continue. This obviously not the only way to do virtual machines, but it is one way if the hardware supports it, to let a virtual machine run very fast as a percentage of the time it is actually running instructions on the hardware. The slowest way to virtualize of course is to virtualize everything including the processor, every instruction in that case would be simulated, this is quite slow but has its own features (virtualizing an arm system on an x86 or x86 on an arm, xyz on an abc, fill in the blanks). And if that is the type of virtual memory you are talking about in an embedded system, well if the embedded system is for the most part indistinguishable from a non-embedded system (an xbox or tivo for example) then well for the same reasons you could allow such a thing. If you were on a microcontroller, well the use cases there would generally mean if you needed more memory you would buy a bigger microcontroller, or add more memory to the system ,or change the needs of the application such that it doesnt need as much memory. there may be exceptions, but it mostly depends on your application and requirements, a general purpose or general purpose like system which allows for applications or their data to be larger than the available ram, will require some sort of solution. the microcontroller in your keyless entry key fob thing or in your tv remote control or clock radio or whatever normally would not have a need to allow "applications" to require more resources than are physically there.
The more important benefit of using virtual memory is that every process gets its own address space which is isolated from every other process's. That way virtual memory helps keep faults contained and improves security and stability. I should note that it is still possible for two processes to share a bit of memory, to facilitate communication (shared mem IPC).
Also you can do other tricks like conserving memory via mapping shared parts into more than one process's (libc comes to mind for embedded use) address space but only having it once in physical mem. Also this gives it a speed boost, you can even enhance it further the way linux does cheapen fork/clone by only copying the in kernel descriptors and leaving the memory image alone up until the first write access is done with a similar idea.
As a last benefit, in modern systems, it's common to do file I/O via mapping the file into the process space (cf. mmap for example).
It's interesting to note that one can get some of the benefits of "virtual memory" without needing a full-fledged MMU. The hardware requirements can sometimes be amazingly light. The PIC 16C505 has a 5-bit address space and 40 bytes of RAM; addresses 0x10 to 0x1F can map to either of two groups of 16 bytes of RAM. When writing an application which needed to manage two different data streams, I arranged so that all the variables associated with one data stream would be in the first group of 16 "switchable" memory locations, and those associated with the other would be at the corresponding addresses in the second group. I could then use the same code to manage both data streams. Simply set the banking bit one way, call the routine, set it the other way, and call the routine again.
One of the reasons Virtual Memory exists is so that your device can multitask. It can also act as your RAM does, thus taking the load off of your physical RAM and swapping the load back and forth.
Let's focus on uniprocessor computer systems. When a process gets created, as far as I know, the page table gets set up which maps the virtual addresses to the physical memory address space. Each process gets its own page table, stored in the kernel address space. But how does the MMU choose the right page table for the process since there is not only one process running and there will be many context switches happening?
Any help is appreciated!
Best,
Simon
Processors have a privileged register called the page table base register (PTBR), on x86 it is CR3. On a context switch, the OS changes the value of the PTBR so that the processor now knows which page table to use. In addition to the PTBR, many modern processors have a notion of an address space number (ASN). Processes are given an address space number (from a limited pool) and this ASN is set in a register on a context switch as well. This ASN is used as part of TLB matching and allows TLB entries from multiple address spaces to coexist. Only when an ASN is reused is it necessary to flush the TLB, and then only for entries matching that ASN. Most x86 implementations are more coarse grained than this and there is a notion of global pages (for shared libraries and shared data).
The MMU in this case is unaware completely of what a process is. The operating system, which keeps tracks of processes, generates a page table for each process, as you say, as they are created. The process for context switching is as follows:
The operating system tells the MMU to use page table located at physical address 0xFOO
The operating system programs the programmable interrupt timer (PIT) to cause a hardware interrupt after BAR milliseconds.
The operating system restores the process state (CPU registers, program counter, etc) and jumps to the correct address.
The process runs until the PIT triggers an interrupt.
The Operating System routine for handling the PIT interrupt then saves the program state (registers etc), uses a scheduling algorithm for determining the next process to run (in a simple case, a circular linked list), then starts over at step 1.
I hope that clears up any doubts you may have. The short answer: The MMU is process agnostic and doesn't know what a process is.
Im interested in understanding how a computer allocates variables for physical memory vs files in virtual memory ( such as on a hard drive ), in terms of how does the computer determine know where to put data. It almost seems random in both memory storage types, but its not because it simply can't put data at a memory address or sector (any location) of a hard drive that's occupied or allocated for another process already. When I was studying how Norton's speed disk ( a program that de-fragments files on hard drives ) on my old W95 system, I noticed from the program's representation of hard drive's data ( a color coded visual map of different data types, e.g. swap files were always first at the top.), consisting of many files spread out all over the hard drive with empty unused areas. In addition some of these areas, I saw what looked like a mix of data and empty space showed a spotty pattern. I want to think its random for that to happen. Like wise, when I was studying the memory addresses of a simple program I wrote in C, I noticed that each version of my program after recompiling it after changes - showed different addresses for segments and offsets. I was expecting the computer to use the same address when I recompiled it. Sometimes the same address would be used, other times it was different. Again, I want to think its random also for memory locations to be chosen by programs. I thought that memory allocation or file writing was based on the first empty space available, written in a contiguous manner.
So my question is, I want to know how and what is it in the logic works of a common computer, that decides where it writes its data in such a arbitrary manner for either type of location (physical RAM or Dynamic )? What area of computer science (if not assembly language) would I need to study that would explains this, almost random behavior?
Thanks in Advance
Something broader and directly from computer science would be a linked list. http://en.wikipedia.org/wiki/Linked_list
Imagine if you had a linked list and simply added items to the end, these items might live linearly in memory or disk or whatever somewhere. But as you remove some items in the middle of the list by having say item number 7 point at item number 9 eliminating item number 8. As with memory allocation for allocs or virtual memory or hard drive sector allocation, etc how fast you fragment your storage has to do with the algorithm you use for allocating the next item.
file systems can/do use a link list type scheme to keep track of what sectors are tied to a single file. it is fast and easy to use the link list but deal with fragmentation. A much slower method would be to have no fragmentation but be constantly copying/moving files around to keep them on linear sectors.
malloc() allocation schemes and MMU allocation schemes also fall under this category. Basically any time you take something, slice it up into fractions and put a virtual interface in front of those fractions to give the appearance to the programmer/user that they are linear. Malloc() (not counting the virtual memory via the MMU) is the other way around allocating a number of linear chunks of those fractions to meed the alloc need, and having an alloc/free scheme that attempts to keep as many large chunks available, just in case, a bad malloc system is one where you have half of your memory free but the maximum malloc that works without an out of memory error is a malloc of a small fraction of that memory, say you have a gig free and can only allocate 4096 bytes.
You should look at virtual memory and TLB (translation lookaside buffer) or paging.
It is not trivial to implement virtual memory and paging. The performance of your whole system depends on it. If it's not done properly your system will thrash.
It is early morning here so Wikipedia will have to do for now: https://en.m.wikipedia.org/wiki/Translation_lookaside_buffer
EDIT:
Those coloured spots you saw in your defrag were chunks on your HDD. Each chunk is of some specified size. Depending on how fragmented your HDD is, you might have portions of your HDD that look like this:
*-*-***-***-*
where * means full, and - means empty
This (above) could be part of one application/file or multiple files; I will assume one file is split across those to simplify my example. At the end of each * there is a pointer to the next location where the next * chunk is (this is called a linked list). The more fragmented your HDD is (or memory) the more of these pointers to next chunk you will have. This in turn uses more space for next pointers instead of using space for data and the result is more overhead when reading that data. If this is a file on disk, you will have multiple seeks (which are bad because they're slow) if your data is not grouped together (locality principle). When you use defrag, it moves and groups all chunks together (as best as it can).
*-*-***-***-*
becomes
*********----
The OS decides paging and virtual memory addressing (and such). TLB is a hardware (a cache) that aids this process (it maps physical memory to virtual memory addresses for fast look up). The CPU communicates with the TLB via MMU
To answer your questions
You should study operating systems.
Yes the locations where to place your files on HDD are decided by the OS. If you deleted a file and download it again, there is no guarantee it will be placed in the same location-most likely not.
A nice summary of all these components and principles I mentioned here work: Click Here. It's a ppt with slides from a Real Time Operating Systems book (if I'm not mistaken the same exact one I used)
I'm working on a project using a LEON2 Processor (Sparc V8).
The processor uses 8Mbytes of RAM that need to be consistency checked during the Self-Test of my Boot.
My issue is that my Boot obviously uses a small part of the RAM for its Heap/BSS/Stack which I cannot modify without crashing my application.
My RAM test is very simple, write a certain value to all the RAM address then read them back to be sure the RAM chip can be addressed.
This method can be used for most of the RAM available but how could I safely check for the consistency of the remaining RAM?
Generally a RAM test that needs to test every single byte will be done as one of the first things that happens when the processor starts. Often the only other thing that's done before it is the hardware initialization that needs to happen for the RAM test to be able to access RAM.
It'll usually be done in assembly language with interrupts disabled, for one reason because that's about the only way you can ensure that no RAM is used.
If you want to perform a RAM test after that point, you still need to do it pretty early in the system start-up. You could maybe do it in two passes - where any variables/stack/whatever the test needs for its own purposes are in low RAM, and that test tests high RAM. Then have the test run again with it's data in high RAM while it tests low RAM.
Another note: verifying that you read back a certain value written is a simple test that maybe better than nothing, but it can miss certain types of common failures (common particularly with external RAM: missing or cross soldered address lines.
You can find more detailed information about basic RAM tests here:
Jack Ganssle, "Testing RAM in Embedded Systems"
Michael Barr, "Fast Accurate Memory Test Suite"
as I am programming a safety-relevant device, I have to do a full RAM test during operation time.
I split the test in two tests:
Addressing test
you write unique values to the addresses reached by each addressing line and after all values are written, the values are read back and compared to the expected values. This test detects short-circuits (or stuck#low/high) of addressing lines (meaning you want to write 0x55 on address 0xFF40, but due to a short-circuit the value is stored at 0xFF80, you cannot detect this by test 2:
Pattern Test:
You save e.g. the first 4 bytes of RAM in CPU's registers and afterwards you first clear the cells, write 0x55, verify, write 0xAA, verify and restore saved content (you can use other patterns of course) and so on. The reason you have to use the registers is that by using a variable, this variable would be destroyed by that test.
You can even test your stack with this test.
In our project, we test 4 cells at a time and we have to run this test until whole RAM is tested.
I hope that helped a bit.
If you do your testing before the C run time environment is up you can trash the Heap and BSS areas without any problems.
Generally the stack does not get used much during run time setup so you may be able trash it with no ill effect. Just check your system.
If you need to use the stack during testing or need to preserve it just move it to an already tested area adjust the stack pointer. After wards just restore the old stack and continue.
There is no easy ways of doing this once you entered your runtime environment.