Peter B. Galvin says that the major factor affecting this decision is the location of the interrupt vector. Since the interrupt vector is usually in low memory, programmers usually place operating system in low memory as well. I really don't understand why the OS has to be next to the interrupt vector.
It sounds like you are in the midst of textbook BS. To begin, you are only talking about parts of the operating system that have physical locations. The rest of the kernel will be in virtual/logical locations.
The interrupt vector is going to be one part of the kernel that has been in physical memory. On most systems you can put the interrupt vector anywhere in memory.
The interrupt vector is part of the operating system. It would make sense to allocate physical memory for the operating system in a contiguous block (ignoring physical memory locations used by devices). Thus, the interrupt vector is likely to be in the same block of reserved physical memory as the rest of the OS uses. The interrupt vector COULD be all alone and separated from the rest of memory used by the operating system but it is easier to do it all in one.
So you're writing your operating system. Where are you going to put this block of reserved memory?
Intuitively, people are going to select either the high end of low end of physical memory.
Related
I want to print the memory location (address) of a variable with:
let x = 1;
println!("{:p}", &x);
This prints the hex value 0x7fff51ef6380 which in decimal is 140734568031104.
My computer has 16GB of RAM, so why this huge number? Does the x64 architecture use a big interval sequence instead of just simple 1 increment for accessing memory location?
In x86, usually the first location starts at 0, then 1, 2, etc. so the highest number you can have is around 4 billion, so the address number was always equals or less than 4 billion.
Why is this not the case with x64?
What you see here is an effect of virtual memory. Memory management is hard and it becomes even harder when the operating system and tens of hundreds of processes have to share the memory. In order to handle this huge complexity, the concept of virtual memory was used. I'll just briefly explain the basics here; the topic is far more complex and you should read about it somewhere else, too.
On most modern computers, each process thinks that it owns (almost) the complete memory space. But processes never deal with physical addresses, but with virtual ones. These virtual addresses are mapped to physical ones each time the process actually reads from memory. This translation of addresses is done by the so called MMU (memory management unit). The rules for how to map the addresses are setup by the operating system.
When you boot your PC, the operating system creates an initial mapping. Every time you start a process, the operating system adds a few slices of physical memory to the process and modifies the mapping appropriately. That way, the process has memory to play with.
On x86_64, the address space is 64 bit wide, so each process thinks it owns all of those 2^64 addresses. This is not true, of course:
There isn't a single PC on the world with that much memory. (In fact, most CPUs today can merely use 280 TB of RAM, since they internally can only use 48bit for addressing physical memory. And even these 280TB are enough for now, apparently.)
Even if you had that much memory, there are other processes which use part of that memory, too.
So what happens when you try to read an address which isn't mapped (which in 64bit land, are the vast majority of the addresses)? The MMU triggers a page fault. This makes the CPU notify the operating system to handle this.
What I mean is that in x86, usually first location starts at 0, then 1, 2, etc. so the highest number you can have is around 4 billion.
That is true, but it is also true if your x86 system has less than 4GB of RAM. Virtual memory exists for quite some time already.
So that's a short summary of why you see such big addresses. Again, please note that I glossed over many details here.
The pointers your program works with are in virtual address space. x86-64 uses 64-bit pointers. This was one of the major goals of AMD64, along with adding more integer and XMM registers. You are correct that i386 only has 32-bit pointers which only cover 4GB of address space in each process.
0x7fff51ef6380 looks like a stack pointer, which I guess makes sense for that code.
Linux on x86-64 (for example) puts the stack near the top of the lower canonical address range: current x86-64 hardware only implements 48-bit virtual addresses and this is the mechanism to prevent software from depending on it. This allows the address space to be extended in the future without breaking software.
The amount of phyiscal RAM in your system has nothing to do with this. You'd see (approximately) the same number on an x86-64 system with 128MB of RAM, +/- stack address space layout randomization (ASLR).
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 8 years ago.
Improve this question
The memory map of a process appears to be fragmented into segments (stack, heap, bss, data, and text),
I was wondering are these segments just an abstraction for the
convenience of the process and the physical RAM is just a linear array
of addresses or is the physical RAM also fragmented into these
segments?
Also if the RAM is not fragmented and is just a linear array then how
does the OS provide the process the abstraction of these segments?
Also how would programming change if the memory map to a process appeared as just a linear array and not divided into segments (with the MMU translating virtual addresses into physical ones)?
In a modern OS supporting virtual memory, it is the address space of the process that is divided into these segments. And in general case that address space of the process is projected onto the physical RAM in a completely random fashion (with some fixed granularity, 4K typically). Address space pages located next to each other do not have to be projected into the neighboring physical pages of RAM. Physical pages of RAM do not have to maintain the same relative order as the process's address space pages. This all means that there is no such separation into segments in RAM and there can't possibly be.
In order to optimize memory access an OS might (and typically will) try to map sequential pages of the process address space to sequential pages in RAM, but that's just an optimization. In general case, the mapping is unpredictable. On top of that the RAM is shared by all processes in the system, with RAM pages belonging to different processes being arbitrarily interleaved in RAM, which eliminates any possibility of having such "segments" in RAM. There's no process-specific ordering or segmentation in RAM. RAM is just a cache for virtual memory mechanism.
Again, every process works with its own virtual address space. This is where these segments can exist. The process has no direct access to RAM. The process doesn't even need to know that RAM exists.
These segments are largely a convenience for the program loader and operating system (though they also provide a basis for coarse-grained protection; execution permission can be limited to text and writes prohibited from rodata).1
The physical memory address space might be fragmented but not for the sake of such application segments. For example, in a NUMA system it might be convenient for hardware to use specific bits to indicate which node owns a given physical address.
For a system using address translation, the OS can somewhat arbitrarily place the segments in physical memory. (With segmented translation, external fragmentation can be a problem; a contiguous range of physical memory addresses may not be available, requiring expensive moving of memory segments. With paged translation, external fragmentation is not a possible. Segmented translation has the advantage of requiring less translation information: each segment requiring only a base and bound with other metadata whereas a memory section would typically have many more than two pages each of which has a base address and metadata.)
Without address translation, placement of segments would necessarily be less arbitrary. Fortunately, most programs do not care about the specific address where segments are placed. (Single address space OSes
(Note that it can be convenient for sharable sections to be in fixed locations. For code this can be used to avoid indirection through a global offset table without requiring binary rewriting in the program loader/dynamic linker. This can also reduce address translation overhead.)
Application-level programming is generally sufficiently abstracted from such segmentation that its existence is not noticeable. However, pure abstractions are naturally unfriendly to intense optimization for physical resource use, including execution time.
In addition, a programming system may choose to use a more complex placement of data (without the application programmer needing to know the implementation details). For example, use of coroutines may encourage using a cactus/spaghetti stack where contiguity is not expected. Similarly, a garbage collecting runtime might provide additional divisions of the address space, not only for nurseries but also for separating leaf objects, which have no references to collectable memory, from non-leaf objects (reducing the overhead of mark/sweep). It is also not especially unusual to provide two stack segments, one for data whose address is not taken (or at least is fixed in size) and one for other data.
1One traditional layout of these segments (with a downward growing stack) in a flat virtual address space for Unix-like OSes places text at the lowest address, rodata immediate above that, initialized data immediately above that, zero-initialized data (bss) immediately above that, heap growing upward from the top of bss, and stack growing downward from the top of the application's portion of the virtual address space.
Having heap and stack growing toward each other allows arbitrary growth of each (for a single thread using that address space!). This placement also allows a program loader to simply copy the program file into memory starting at the lowest address, groups memory by permission, and can sometimes allow a single global pointer to address all of the global/static data range (rodata, data, and bss).
The memory map to a process appears fragmented into segments (stack, heap, bss, data, and text)
That's the basic mapping used by Unix; other operating systems use different schemes. Generally, though, they split the process memory space into separate segments for executing code, stack, data, and heap data.
I was wondering are these segments are just abstraction for the processes for convience and the physical RAM is just a linear array of addresses or the physical RAM is also fragmented into these segments?
Depends.
Yes, these segments are created and managed by the OS for the benefit of the process. But physical memory can be arranged as linear addresses, or banked segments, or non-contiguous blocks of RAM. It's up to the OS to manage the total system memory space so that each process can access its own portion of it.
Virtual memory adds yet another layer of abstraction, so that what looks like linear memory locations are in fact mapped to separate pages of RAM, which could be anywhere in the physical address space.
Also if the RAM is not fragmanted and is just a linear array then how the OS provides the process the abstraction of these segments?
The OS manages all of this by using virtual memory mapping hardware. Each process sees contiguous memory areas for its code, data, stack, and heap segments. But in reality, the OS maps the pages within each of these segments to physical pages of RAM. So two identical running processes will see the same virtual address space composed of contiguous memory segments, but the memory pages comprising these segments will be mapped to entirely different physical RAM pages.
But bear in mind that physical RAM may not actually be one contiguous block of memory, but may in fact be split across multiple non-adjacent blocks or memory banks. It is up to the OS to manage all of this in a way that is transparent to the processes.
Also how the programming would change if the memory map to a process would appear just as a linear array and not divided into segments?, and then the MMU would just translate these virtual addresses into physical ones.
The MMU always operates that way, translating virtual memory addresses into physical memory addresses. The OS sets up and manages the mapping of each page of each segment for each process. Each time the process exceeds its stack allocation, for example, the OS traps a segment fault and adds another page to the process's stack segment, mapping the virtual page to a physical page selected from available memory.
Virtual memory also allows the OS to swap out process pages temporarily to disk, so that the total amount of virtual memory occupied by all of the running processes can easily exceed the actual physical memory RAM space of a system. Only the currently active executing processes actually have access to real physical RAM pages.
I was wondering are these segments are just abstraction for the
processes for convience and the physical RAM is just a linear array of
addresses or the physical RAM is also fragmented into these segments?
This in fact highly depends on architecture. Some will have hardware tools (e.g. descriptor registers for x86) to split the RAM into segments. Others just keep this information in software (OS kernel information for this process). Also some segments information are totally irrelevant on execution, they're used merely for code/data loading (e.g. relocation segments).
Also if the RAM is not fragmanted and is just a linear array then how
the OS provides the process the abstraction of these segments?
Process code never references to segments, he only knows about addresses, so the OS has nothing to abstract.
Also how the programming would change if the memory map to a process
would appear just as a linear array and not divided into segments?,
and then the MMU would just translate these virtual addresses into
physical ones
Programming would not be affected. When you program in C you don't define any of these segments, and code also doesn't reference these segments. These segments are to keep an ordered layout, and don't even need to be the same across OS.
So my understanding is that every process has its own virtual memory space ranging from 0x0 to 0xFF....F. These virtual addresses correspond to addresses in physical memory (RAM). Why is this level of abstraction helpful? Why not just use the direct addresses?
I understand why paging is beneficial, but not virtual memory.
There are many reasons to do this:
If you have a compiled binary, each function has a fixed address in memory and the assembly instructions to call functions have that address hardcoded. If virtual memory didn't exist, two programs couldn't be loaded into memory and run at the same time, because they'd potentially need to have different functions at the same physical address.
If two or more programs are running at the same time (or are being context-switched between) and use direct addresses, a memory error in one program (for example, reading a bad pointer) could destroy memory being used by the other process, taking down multiple programs due to a single crash.
On a similar note, there's a security issue where a process could read sensitive data in another program by guessing what physical address it would be located at and just reading it directly.
If you try to combat the two above issues by paging out all the memory for one process when switching to a second process, you incur a massive performance hit because you might have to page out all of memory.
Depending on the hardware, some memory addresses might be reserved for physical devices (for example, video RAM, external devices, etc.) If programs are compiled without knowing that those addresses are significant, they might physically break plugged-in devices by reading and writing to their memory. Worse, if that memory is read-only or write-only, the program might write bits to an address expecting them to stay there and then read back different values.
Hope this helps!
Short answer: Program code and data required for execution of a process must reside in main memory to be executed, but main memory may not be large enough to accommodate the needs of an entire process.
Two proposals
(1) Using a very large main memory to alleviate any need for storage allocation: it's not feasible due to very high cost.
(2) Virtual memory: It allows processes that may not be entirely in the memory to execute by means of automatic storage allocation upon request. The term virtual memory refers to the abstraction of separating LOGICAL memory--memory as seen by the process--from PHYSICAL memory--memory as seen by the processor. Because of this separation, the programmer needs to be aware of only the logical memory space while the operating system maintains two or more levels of physical memory space.
More:
Early computer programmers divided programs into sections that were transferred into main memory for a period of processing time. As higher level languages became popular, the efficiency of complex programs suffered from poor overlay systems. The problem of storage allocation became more complex.
Two theories for solving the problem of inefficient memory management emerged -- static and dynamic allocation. Static allocation assumes that the availability of memory resources and the memory reference string of a program can be predicted. Dynamic allocation relies on memory usage increasing and decreasing with actual program needs, not on predicting memory needs.
Program objectives and machine advancements in the '60s made the predictions required for static allocation difficult, if not impossible. Therefore, the dynamic allocation solution was generally accepted, but opinions about implementation were still divided.
One group believed the programmer should continue to be responsible for storage allocation, which would be accomplished by system calls to allocate or deallocate memory. The second group supported automatic storage allocation performed by the operating system, because of increasing complexity of storage allocation and emerging importance of multiprogramming.
In 1961, two groups proposed a one-level memory store. One proposal called for a very large main memory to alleviate any need for storage allocation. This solution was not possible due to very high cost. The second proposal is known as virtual memory.
cne/modules/vm/green/defn.html
To execute a process its data is needed in the main memory (RAM). This might not be possible if the process is large.
Virtual memory provides an idealized abstraction of the physical memory which creates the illusion of a larger virtual memory than the physical memory.
Virtual memory combines active RAM and inactive memory on disk to form
a large range of virtual contiguous addresses. implementations usually require hardware support, typically in the form of a memory management
unit built into the CPU.
The main purpose of virtual memory is multi-tasking and running large programmes. It would be great to use physical memory, because it would be a lot faster, but RAM memory is a lot more expensive than ROM.
Good luck!
In modern-day operating systems, memory is available as an abstracted resource. A process is exposed to a virtual address space (which is independent from address space of all other processes) and a whole mechanism exists for mapping any virtual address to some actual physical address.
My doubt is:
If each process has its own address space, then it should be free to access any address in the same. So apart from permission restricted sections like that of .data, .bss, .text etc, one should be free to change value at any address. But this usually gives segmentation fault, why?
For acquiring the dynamic memory, we need to do a malloc. If the whole virtual space is made available to a process, then why can't it directly access it?
Different runs of a program results in different addresses for variables (both on stack and heap). Why is it so, when the environments for each run is same? Does it not affect the amount of addressable memory available for usage? (Does it have something to do with address space randomization?)
Some links on memory allocation (e.g. in heap).
The data available at different places is very confusing, as they talk about old and modern times, often not distinguishing between them. It would be helpful if someone could clarify the doubts while keeping modern systems in mind, say Linux.
Thanks.
Technically, the operating system is able to allocate any memory page on access, but there are important reasons why it shouldn't or can't:
different memory regions serve different purposes.
code. It can be read and executed, but shouldn't be written to.
literals (strings, const arrays). This memory is read-only and should be.
the heap. It can be read and written, but not executed.
the thread stack. There is no reason for two threads to access each other's stack, so the OS might as well forbid that. Moreover, the tread stack can be de-allocated when the tread ends.
memory-mapped files. Any changes to this region should affect a specific file. If the file is open for reading, the same memory page may be shared between processes because it's read-only.
the kernel space. Normally the application should not (or can not) access that region - only kernel code can. It's basically a scratch space for the kernel and it's shared between processes. The network buffer may reside there, so that it's always available for writes, no matter when the packet arrives.
...
The OS might assume that all unrecognised memory access is an attempt to allocate more heap space, but:
if an application touches the kernel memory from user code, it must be killed. On 32-bit Windows, all memory above 1<<31 (top bit set) or above 3<<30 (top two bits set) is kernel memory. You should not assume any unallocated memory region is in the user space.
if an application thinks about using a memory region but doesn't tell the OS, the OS may allocate something else to that memory (OS: sure, your file is at 0x12341234; App: but I wanted to store my data there). You could tell the OS by touching the end of your array (which is unreliable anyways), but it's easier to just call an OS function. It's just a good idea that the function call is "give me 10MB of heap", not "give me 10MB of heap starting at 0x12345678"
If the application allocates memory by using it then it typically does not de-allocate at all. This can be problematic as the OS still has to hold the unused pages (but the Java Virtual Machine does not de-allocate either, so hey).
Different runs of a program results in different addresses for variables
This is called memory layout randomisation and is used, alongside of proper permissions (stack space is not executable), to make buffer overflow attacks much more difficult. You can still kill the app, but not execute arbitrary code.
Some links on memory allocation (e.g. in heap).
Do you mean, what algorithm the allocator uses? The easiest algorithm is to always allocate at the soonest available position and link from each memory block to the next and store the flag if it's a free block or used block. More advanced algorithms always allocate blocks at the size of a power of two or a multiple of some fixed size to prevent memory fragmentation (lots of small free blocks) or link the blocks in a different structures to find a free block of sufficient size faster.
An even simpler approach is to never de-allocate and just point to the first (and only) free block and holds its size. If the remaining space is too small, throw it away and ask the OS for a new one.
There's nothing magical about memory allocators. All they do is to:
ask the OS for a large region and
partition it to smaller chunks
without
wasting too much space or
taking too long.
Anyways, the Wikipedia article about memory allocation is http://en.wikipedia.org/wiki/Memory_management .
One interesting algorithm is called "(binary) buddy blocks". It holds several pools of a power-of-two size and splits them recursively into smaller regions. Each region is then either fully allocated, fully free or split in two regions (buddies) that are not both fully free. If it's split, then one byte suffices to hold the size of the largest free block within this block.
Can anybody tell me a unix command that can be used to find the number of memory accesses that took place in a given interval. vmstat, top and sar only give the amount of physical memory space occupied/available .. But do not give the number of memory of accesses in a given interval
If I understand what you're asking, such a feature would almost certainly require hardware support at a very low level (e.g. a counter of some sort that monitors memory bus activity).
I don't think such support is available for the common architectures supported by
Unix or Linux, so I'm going to go out on a limb and say that no such Unix command exists.
The situation is somewhat different when considering memory in units of pages,
because most architectures that support virtual memory have dedicated MMU hardware
which operates at that level of granularity, and can be accessed by the operating
system. But as far as I know, the sorts of counter data you'd get from the MMU would
represent events like page faults, allocations, and releases, rather than individual
reads or writes.