I have few doubts regarding the memory management in a x86_64 Linux Operating System.
If I allocate an array of 2000 bytes (statically - arr[2000]; or dynamically - malloc(2000);) from my user space code, are these going to be a contiguous memory in physical memory?
If I allocate memory (same above example, statically - arr[2000]; or dynamically - malloc(2000);) will there be a page table updation to map to these 2000 bytes in physical memory, so that the future references to these memory addresses can be found from the Page Table Entry?
1) Very unlikely. Its possible that "your" malloc() might seem to produce the result but you cannot rely on it.
What you would want to do is malloc(4000) and then have two pointers. One to the malloc and the other at pointer1+2000.
Be careful that when you free(pointer1) that you also nullify pointer2.
2) Not until you reference a byte within the area.
Related
I'm developing a C-library to be used under uCOS-III. The CPU is an ARM Cortex M4 SAM4C. Within the library I want to use a third party product X, whose particular name is not relevant here. The source code for X is completely available and compiles without problems.
Inside X a lot of memory allocations are executed, using calloc() and free().
The problem is, that plain usage of malloc is not advisable for embedded systems, because of memory fragmentation. The documentation for uCOS-III explicitly advises against using malloc - instead OSMemCreate/OSMemGet/OSMemPut shall be used to allocate and free chunks of memory out of a statically allocated memory block.
Question-1:
What is the general advice to get around the "standard implementation" of malloc? I would prefer a kind of malloc, where I have access to a fixed memory pool (e.g. dedicated for a special task)
Question-2:
How should OSMemCreate() be used correctly? I have first to initialize a memory partition with a certain block size. The amount of requested memory is between 4 Bytes and about 800 bytes. I can get blocks on request, but with fixed size. If block-size=4 I cannot allocate 16 Bytes, since blocks are not contiguous in memory. If block-size=800 and I need only 4 bytes, most of the block is left unused and I will very soon run out of blocks.
So I don't know, how to solve my original problem by use of OSMemCreate...
Can anybody give me an advice how I could proceed?
Many thanks,
Michael
1) Don't link with the standard library version of malloc/free. Instead create your own implementation of malloc/free that serves as a wrapper to OSMemGet/OSMemPut.
2) You can create more than one memory partition with OSMemCreate. Create small, medium, and large partitions that hold block sizes which are tuned for your application to reduce waste.
If you want malloc to get an appropriately sized block from your various memory partitions then you'll have to invent some magic so that free returns the block to the appropriate memory partition. (Perhaps malloc allocates an extra word, stores the pointer to the memory partition in the first word, and then returns the address after the word where the pointer is stored. Then free knows to get the memory partition pointer from the preceding word.)
An alternative to using malloc/free is to rewrite that code to use statically allocated variables or call OSMemGet/OSMemPut directly.
In linux if malloc can't allocate data chunk from preallocated page. it uses mmap() or brk() to assign new pages. I wanted to clarify a few things :
I don't understand the following statment , I thought that when I use mmap() or brk() the kernel maps me a whole page. but the allocator alocates me only what I asked for from that new page? In this case trying to access unallocated space (In the newly mapped page) will not cause a page fault? (I understand that its not recommended )
The libc allocator manages each page: slicing them into smaller blocks, assigning them to processes, freeing them, and so on. For example, if your program uses 4097 bytes total, you need to use two pages, even though in reality the allocator gives you somewhere between 4105 to 4109 bytes
How does the allocator knows the VMA bounders ?(I assume no system call used) because the VMA struct that hold that information can only get accessed from kernel mode?
The system-level memory allocation (via mmap and brk) is all page-aligned and page-sized. This means if you use malloc (or other libc API that allocates memory) some small amount of memory (say 10 bytes), you are guaranteed that all the other bytes on that page of memory are readable without triggering a page fault.
Malloc and family do their own bookkeeping within the pages returned from the OS, so the mmap'd pages used by libc also contain a bunch of malloc metadata, in addition to whatever space you've allocated.
The libc allocator knows where everything is because it invokes the brk() and mmap() calls. If it invokes mmap() it passes in a size, and the kernel returns a start address. Then the libc allocator just stores those values in its metadata.
Doug Lea's malloc implementation is a very, very well documented memory allocator implementation and its comments will shed a lot of light on how allocators work in general:
http://g.oswego.edu/dl/html/malloc.html
Suppose there is 50 MB to allocate and there is this for loop which allocates memory in each iteration. What happens when the following for loop runs.
for(int i=0; i < 20; i++)
{
int *p = malloc(5MB);
}
I was asked this question in an interview. Can someone guide me on this and direct me to the necessary topics which I must learn in order to understand such situations.
If this is a system utilising virtual memory then the situation is obviously more complex than malloc simply failing and returning null pointers. In this case the malloc call will result in allocation of pages of memory in the virtual address space. When this memory is accessed a page fault will result, control will be given to the os memory manager, and this will map the virtual memory pages to pages of physical memory. When the physical memory available is full then the memory manager will generally handle further page faults by writing data which is currently in physical memory into disk backing (or simply discarding this data if it is already backed by a disk file), and then remapping this now available physical memory for the virtual memory page which originally resulted in the page fault. Any attempts to access virtual address pages which have previously been written out to disk backing will result in a similar process occurring.
Wikipedia contains a reasonably basic overview of this process (http://en.wikipedia.org/wiki/Paging), including some implementation details for different os's. More detailed information is available from a number of other sources, eg. the intel architectures software development manuals (http://www.intel.com/content/www/us/en/processors/architectures-software-developer-manuals.html)
According to documentation for malloc (eg. http://www.cplusplus.com/reference/cstdlib/malloc/) you will get a pointer to a memory block of 5MB (noting that the argument for malloc should be the number of bytes required, not MB..) the first 10 times (maybe 9 times if there is overhead associated with the malloc'ed space which is taken out of the 50MB available), and a pointer to this space will be returned. Following this further 5MB chunks of memory will not be available, and malloc will fail, returning a null pointer.
In almost all books and articles I have read about about HIGHMEM in Linux kernel, they say while using 3:1 split, not all of the 1GB is available to the kernel for mapping. And normally its 896MB or so, with the rest being used for kernel data structures, memory maps, page tables and such.
My question is, what exactly are these data structures? Page tables are normally accessed via a page table address register, right? And the base address of page table is normally stored as a physical address. Now why do one need to reserve a virtual address space for the entire table?
Similarly, I read about kernel code itself occupying space. What does that have to do with virtual address space? Is it not the physical memory that would be consumed for storing the code?
And finally, these data structures why do they have to reserve the 128MB space? Why can't they be used out of the entire 1GB address space, as required, like any other normal data structure in kernel would do?
I have gone through LDD3, Professional Linux Kernel Architecture and several posts here at stack-overflow (like: Why Linux Kernel ZONE_NORMAL is limited to 896 MB?) and an older LWN article, but found no specific information on the same.
With regards to page tables, it's true that the MMU wouldn't care if the page tables themselves weren't mapped in the virtual address space - for the purposes of address translations, that would be OK. But when the kernel needs to modify the page tables, they do need to be mapped in the virtual address space - and the kernel can't just map them in "just in time", because it needs to modify the page tables themselves to do that. It's a chicken-and-egg problem, which means that the page tables need to remain mapped at all times.
A similar problem exists with the kernel code. For the code to execute, it must be mapped in the virtual address space - and if the code that does the page table modification were itself not present, we'd have a similar chicken-and-egg problem. Given this, it's easier to leave the entireity of the kernel code mapped all the time, along with the kernel-mode stacks and any kernel data structures access by code where you wouldn't want to potentially take a page fault. One large example of such data structures is the array of struct page structures, representing each physical memory page.
The 128MB reserve is not for a specific data structure, that always use it.
It's virtual memory, reserved for various users, which might use it. Normally, it's not all used.
About physical and virtual memory: every allocation needs three things - a physical page, a virutal page, and mapping connecting the two. Linux almost never uses physical addresses directly, it always passes virtual address translation.
For most kernel memory allocation (called lowmem), this translation is quite simple - subtract some constant from the virtual address to get the physical. But still, a virtual address is used.
Linux's memory management was written when the virtual memory space (4GB) was much larger
than the physical memory, even on the largest machines. In such cases, wasting virtual addresses is not a problem. Today, when physical memory is large, this leads to inefficiencies and problems.
The vmalloc virtual address range is used by any caller of vmalloc. For example:
1. Loading kernel drivers (using modprobe or insmod).
2. Kernel modules often allocate with vmalloc. The alternative function kmalloc used to be limited to 128K, and it rounds the size up to a power of 2, so vmalloc is often preferred for large allocations.
When a process requests physical memory pages from the Linux kernel, the kernel does its best to provide a block of pages that are physically contiguous in memory. I was wondering why it matters that the pages are PHYSICALLY contiguous; after all, the kernel can obscure this fact by simply providing pages that are VIRTUALLY contiguous.
Yet the kernel certainly tries its hardest to provide pages that are PHYSICALLY contiguous, so I'm trying to figure out why physical contiguity matters so much. I did some research &, across a few sources, uncovered the following reasons:
1) makes better use of the cache & achieves lower avg memory access times (GigaQuantum: I don’t understand: how?)
2) you have to fiddle with the kernel page tables in order to map pages that AREN’T physically contiguous (GigaQuantum: I don’t understand this one: isn’t each page mapped separately? What fiddling has to be done?)
3) mapping pages that aren’t physically contiguous leads to greater TLB thrashing (GigaQuantum: I don’t understand: how?)
Per the comments I inserted, I don't really understand these 3 reasons. Nor did any of my research sources adequately explain/justify these 3 reasons. Can anyone explain these in a little more detail?
Thanks! Will help me to better understand the kernel...
The main answer really lies in your second point. Typically, when memory is allocated within the kernel, it isn't mapped at allocation time - instead, the kernel maps as much physical memory as it can up-front, using a simple linear mapping. At allocation time it just carves out some of this memory for the allocation - since the mapping isn't changed, it has to already be contiguous.
The large, linear mapping of physical memory is efficient: both because large pages can be used for it (which take up less space for page table entries and less TLB entries), and because altering the page tables is a slow process (so you want to avoid doing this at allocation/deallocation time).
Allocations that are only logically linear can be requested, using the vmalloc() interface rather than kmalloc().
On 64 bit systems the kernel's mapping can encompass the entireity of physical memory - on 32 bit systems (except those with a small amount of physical memory), only a proportion of physical memory is directly mapped.
Actually the behavior of memory allocation you describe is common for many OS kernels and the main reason is kernel physical pages allocator. Typically, kernel has one physical pages allocator that is used for allocation of pages for both kernel space (including pages for DMA) and user space. In kernel space you need continuos memory, because it's expensive (for in-kernel code) to map pages every time you need them. On x86_64, for example, it's completely worthless because kernel can see the whole address space (on 32bit systems there's 4G limitation of virtual address space, so typically top 1G are dedicated to kernel and bottom 3G to user-space).
Linux kernel uses buddy algorithm for page allocation, so that allocation of bigger chunk takes fewer iterations than allocation of smaller chunk (well, smaller chunks are obtained by splitting bigger chunks). Moreover, using of one allocator for both kernel space and user space allows the kernel to reduce fragmentation. Imagine that you allocate pages for user space by 1 page per iteration. If user space needs N pages, you make N iterations. What happens if kernel wants some continuos memory then? How can it build big enough continuos chunk if you stole 1 page from each big chunk and gave them to user space?
[update]
Actually, kernel allocates continuos blocks of memory for user space not as frequently as you might think. Sure, it allocates them when it builds ELF image of a file, when it creates readahead when user process reads a file, it creates them for IPC operations (pipe, socket buffers) or when user passes MAP_POPULATE flag to mmap syscall. But typically kernel uses "lazy" page loading scheme. It gives continuos space of virtual memory to user-space (when user does malloc first time or does mmap), but it doesn't fill the space with physical pages. It allocates pages only when page fault occurs. The same is true when user process does fork. In this case child process will have "read-only" address space. When child modifies some data, page fault occurs and kernel replaces the page in child address space with a new one (so that parent and child have different pages now). Typically kernel allocates only one page in these cases.
Of course there's a big question of memory fragmentation. Kernel space always needs continuos memory. If kernel would allocate pages for user-space from "random" physical locations, it'd be much more hard to get big chunk of continuos memory in kernel after some time (for example after a week of system uptime). Memory would be too fragmented in this case.
To solve this problem kernel uses "readahead" scheme. When page fault occurs in an address space of some process, kernel allocates and maps more than one page (because there's possibility that process will read/write data from the next page). And of course it uses physically continuos block of memory (if possible) in this case. Just to reduce potential fragmentation.
A couple of that I can think of:
DMA hardware often accesses memory in terms of physical addresses. If you have multiple pages worth of data to transfer from hardware, you're going to need a contiguous chunk of physical memory to do so. Some older DMA controllers even require that memory to be located at low physical addresses.
It allows the OS to leverage large pages. Some memory management units allow you to use a larger page size in your page table entries. This allows you to use fewer page table entries (and TLB slots) to access the same quantity of virtual memory. This reduces the likelihood of a TLB miss. Of course, if you want to allocate a 4MB page, you're going to need 4MB of contiguous physical memory to back it.
Memory-mapped I/O. Some devices could be mapped to I/O ranges that require a contiguous range of memory that spans multiple frames.
Contiguous or Non-Contiguous Memory Allocation request from the kernel depends on your application.
E.g. of Contiguous memory allocation: If you require a DMA operation to be performed then you will be requesting the contiguous memory through kmalloc() call as DMA operation requires a memory which is also physically contiguous , as in DMA you will provide only the starting address of the memory chunk and the other device will read or write from that location.
Some of the operation do not require the contiguous memory so you can request a memory chunk through vmalloc() which gives the pointer to non contagious physical memory.
So it is entirely dependent on the application which is requesting the memory.
Please remember that it is a good practice that if you are requesting the contiguous memory than it should be need based only as kernel is trying best to allocation the memory which is physically contiguous.Well kmalloc() and vmalloc() has their limits also.
Placing things we are going to be reading a lot physically close together takes advantage of spacial locality, things we need are more likely to be cached.
Not sure about this one
I believe this means if pages are not contiguous, the TLB has to do more work to find out where they all are. If they are contigous, we can express all the pages for a processes as PAGES_START + PAGE_OFFSET. If they aren't, we need to store a seperate index for all of the pages of a given processes. Because the TLB has a finite size and we need to access more data, this means we will be swapping in and out a lot more.
kernel does not need physically contiguous pages actually it just needs efficencies ans stabilities.
monolithic kernel tends to have one page table for kernel space shared among processes
and does not want page faults on kernel space that makes kernel designs too complex
so usual implementations on 32 bit architecture is always 3g/1g split for 4g address space
for 1g kernel space, normal mappings of code and data should not generate recursive page faults that is too complex to manage:
you need to find empty page frames, create mapping on mmu, and handle tlb flush for new mappings on every kernel side page fault
kernel is already busy of doing user side page faults
furthermore, 1:1 linear mapping could have much less page table entries because it can utilize bigger size of page unit (>4kb)
less entries leads to less tlb misses.
so buddy allocator on kernel linear address space always provides physically contiguous page frames
even most codes doesn't need contiguous frames
but many device drivers which need contiguous page frames already believe that allocated buffers through general kernel allocator are physically contiguous