Generally as we know virtual memory is larger than physical memory.But when is it advantageous to define virtual memory smaller than physical memory?
If you have pointer-heavy code, you can save memory by choosing a smaller address space. For example, a pointer on a 32-bit platform occupies 4 bytes versus 8 bytes on 64-bit. The same goes for integer types like size_t.
This only works and makes sense if:
Your code/application/server uses multiple processes and all processes together need more memory than the amount of virtual memory (otherwise you wouldn't need more physical than virtual memory).
Your platform supports more physical than virtual memory (for example, Intel PAE).
The smaller amount of virtual memory is enough for each single process.
Imagine a large server system supporting multiple users. You don't want users to hog memory, so you restrict the size of the logical (virtual) address space by limiting page table size.
Related
A comment in this blog states:
We know how to make chunked heaps, but there would be some overhead to
using them. We have more requests for faster storage management than
we do for larger heaps in the 32-bit JVM. If you really want large
heaps, switch to the 64-bit JVM. We still need contiguous memory,
but it's much easier to get in a 64-bit address space.
This implication of the above statement is that it is easier to get contiguous memory in a 64-bit address space. Is this true? If so why?
That's very true. A process must allocate memory from the virtual memory address space. Which stores both code and data and whose size is restricted by the addressing capability of the architecture. You can never address more than 2^32 bytes in a 32-bit process, not counting bank-switching tricks. That's 4 gigabytes. The operating system typically takes a big chunk out of that as well, on 32-bit Windows for example that cuts down the addressable VM size to 2 gigabytes.
Ideally, allocations are made so that they fit snugly together. That very rarely works out in practice. Shared libraries or DLLs in particular need to pick a preferred load address and that has to be guessed up front when the library is built.
So in practice, the allocations are made from the holes in between existing ones and the largest possible contiguous allocation you can get is restricted by the size of the largest hole. Usually much smaller than the addressable VM size, on Windows it is typically around 650 megabytes. That tends to go down-hill from there as the available address space is getting fragmented by allocations. Particularly by native code that can't afford to have allocations moved by a compacting garbage collector. If you use Windows then you can get insight in the VM allocations with the SysInternals' VMMap utility.
This problem completely disappears in a 64-bit process. The theoretical addressable virtual memory size is 2^64, an enormous number. So large that current processors don't implement it, they can go up to 2^48. Further restricted by the operating system version you have and its willingness to keep page mapping tables for that much VM. Eight terabytes is a typical limit. By implication, the holes between allocations are huge. Your program will keel over on paging file thrashing before it dies from OOM.
I can't speak for how the JVM is implemented obviously, but from a purely theoretical viewpoint, if you have a significantly larger virtual address space (eg 64-bit as compared with 32-bit) it should be significantly easier to find a large block of contiguous memory which is available for allocation (going to extremes - you've got no chance of finding a contiguous 4GB of free memory in a 32-bit address space, but a significant chance of finding this space in a full 64-bit address space).
It should be noted that whatever the virtual address space size, this is still going to be implemented by allocation of (probably) non-contiguous physical memory pages, particularly if the requested allocation is large - the larger virtual address space just means there are a likely to be a lot more contiguous virtual addresses available for use.
Being asked this in my homework.
I don't know how to proceed.
Without virtual memory, you couldn't have something as simple as fork without having to swap processes in and out on every task switch (because they'd be using the same physical memory addresses). You also couldn't memory map files or fault-load executables. So yes, virtual memory would still be useful even if not being used to permit more virtual memory than physical memory.
I trying to to make the linux memory management a little bit more clear for tuning and performances purposes.
By reading this very interesting redbook "Linux Performance and Tuning Guidelines" found on the IBM website I came across something I don't fully understand.
On 32-bit architectures such as the IA-32, the Linux kernel can directly address only the first gigabyte of physical memory (896 MB when considering the reserved range). Memory above the so-called ZONE_NORMAL must be mapped into the lower 1 GB. This mapping is completely transparent to applications, but allocating a memory page in ZONE_HIGHMEM causes a small performance degradation.
why the memory above 896 MB has to be mapped into the lower 1GB ?
Why there is an impact on performances by allocating a memory page in ZONE_HIGHMEM ?
what is the ZONE_HIGHMEM used for then ?
why a kernel that is able to recognize up to 4gb ( CONFIG_HIGHMEM=y ) can just use the first gigabyte ?
Thanks in advance
When a user process traps in to the kernel, the page tables are not changed. This means that one linear address space must be able to cover both the memory addresses available to the user process, and the memory addresses available to the kernel.
On IA-32, which allows a 4GB linear address space, usually the first 3GB of the linear address space are allocated to the user process, and the last 1GB of the linear address space is allocated to the kernel.
The kernel must use its 1GB range of addresses to be able to address any part of physical memory it needs to. Memory above 896MB is not "mapped into the low 1GB" - what happens is that physical memory below 896MB is assigned a permanent linear address in the kernel's part of the linear address space, whereas as memory above that limit must be assigned a temporary mapping in the remaining part of the linear address space.
There is no impact on performance when mapping a ZONE_HIGHMEM page into a userspace process - for a userspace process, all physical memory pages are equal. The impact on performance exists when the kernel needs to access a non-user page in ZONE_HIGHMEM - to do so, it must map it into the linear address space if it is not already mapped.
I've read that, on a 32-bit system with 4GB system memory, 2GB is allocated to user mode and 2GB allocated to kernel mode. But, If I had a system with 512 MB of memory, would it be partitioned as 256 MB to user and 256 MB to kernel address space?
You are confusing physical and virtual memory. 2GB is allocated to user/system, but it is virtual memory. It is even more correct to say that they are not rather allocated but they constitute an addressing space. Initially this space is not bound to physical memory at all. When application actually needs memory (first time is at start up) physical memory is allocated and some addresses from address space are mapped to it. When memory is allocated but not used long enough or PC is running out of physical memory data can be dumped in swap file, and stay there until requested. This mapping is transparent for application and it has no idea where data currently is: on chip or on HDD. So the address space is always splitted the same way.
This is not about memory (physical or virtual), but about address space.
You can plug 16GB of physical memory into your computer and make a 100GB swapfile, but 32-bit (non-enterprise) Windows will still only see 4GB (and subtract 0.75 GB for GPU memory and such). Via PAE, it could use more, but non-enterprise versions won't do that.
On top of the actual amount of memory, there is address space, which is limited to 4GB as well. Basically it is no more and no less than the collection of "numbers" (which, in this case, are addresses) that can be represented by a 32 bit number.
Since the kernel will need memory too, there is some arbitrary line drawn, which happens to be at the 2GB boundary for 32bit Windows, but can be configured differently, too.
It has nothing to do with the amount of memory on your computer (virtual or phsyical), but it is a limiting factor of how much memory you can use within a single program instance. It is not, however, a limiting factor on the memory that several programs could use.
As far as I can tell, what you are referring to are limits of how much memory can be allocated. This is much different than how much memory the OS allocated during runtime.
I am playing with an MSDN sample to do memory stress testing (see: http://msdn.microsoft.com/en-us/magazine/cc163613.aspx) and an extension of that tool that specifically eats physical memory (see http://www.donationcoder.com/Forums/bb/index.php?topic=14895.0;prev_next=next). I am obviously confused though on the differences between Virtual and Physical Memory. I thought each process has 2 GB of virtual memory (although I also read 1.5 GB because of "overhead". My understanding was that some/all/none of this virtual memory could be physical memory, and the amount of physical memory used by a process could change over time (memory could be swapped out to disc, etc.)I further thought that, in general, when you allocate memory, the operating system could use physical memory or virtual memory. From this, I conclude that dwAvailVirtual should always be equal to or greater than dwAvailPhys in the call GlobalMemoryStatus. However, I often (always?) see the opposite. What am I missing.
I apologize in advance if my question is not well formed. I'm still trying to get my head around the whole memory management system in Windows. Tutorials/Explanations/Book recs are most welcome!
Andrew
That was only true in the olden days, back when RAM was expensive. The operating system maps pages of virtual memory to RAM as needed. If there isn't enough RAM to satisfy a program's request, it starts unmapping pages to make room. If such a page contains data instead of code, it gets written to the paging file. Whenever the program accesses that page again, it generates a paging fault, letting the operating system read the page back from disk.
If the machine has little RAM and lots of processes consuming virtual memory pages, that can cause a very unpleasant effect called "thrashing". The operating system is constantly accessing the disk and machine performance slows down to a crawl.
More RAM means less disk access. There's very little reason not to use 3 or 4 GB of RAM on a 32-bit operating system, it's cheap. Even if you do not get to use all 4 GB, not all of it will be addressable due hardware devices taking space on the address bus (video, mostly). But that won't change the size of the virtual memory accessible by user code, it is still 2 Gigabytes.
Windows Internals is a good book.
The amount of virtual memory is limited by size of the address space - which is 4GB per process on a 32-bit system. And you have to subtract from this the size of regions reserved for system use and the amount of VM used already by your process (including all the libraries mapped to its address space).
On the other hand, the total amount of physical memory may be higher than the amount of virtual memory space the system has left free for your process to use (and these days it often is).
This means that if you have more than ~2GB or RAM, you can't use all your physical memory in one process (since there's not enough virtual memory space to map it to), but it can be used by many processes. Note that this limitation is removed in a 64-bit system.
I don't know if this is your issue, but the MSDN page for the GlobalMemoryStatus function contains the following warning:
On computers with more than 4 GB of memory, the GlobalMemoryStatus function can return incorrect information, reporting a value of –1 to indicate an overflow. For this reason, applications should use the GlobalMemoryStatusEx function instead.
Additionally, that page says:
On Intel x86 computers with more than 2 GB and less than 4 GB of memory, the GlobalMemoryStatus function will always return 2 GB in the dwTotalPhys member of the MEMORYSTATUS structure. Similarly, if the total available memory is between 2 and 4 GB, the dwAvailPhys member of the MEMORYSTATUS structure will be rounded down to 2 GB. If the executable is linked using the /LARGEADDRESSAWARE linker option, then the GlobalMemoryStatus function will return the correct amount of physical memory in both members.
Since you're referring to members like dwAvailPhys instead of ullAvailPhys, it sounds like you're using a MEMORYSTATUS structure instead of a MEMORYSTATUSEX structure. I don't know the consequences of that on a 64-bit platform, but on a 32-bit platform that definitely could cause incorrect memory sizes to be reported.