I'm sharing many memory blocks between two processes in Linux, one process receives data from network, and the other uses the data in those blocks (buffers).
Some buffers are small (hundreds bytes), and some buffers are very large(2G~4G bytes), while they all have same struct (array of a struct) and will be treated with a same algorithm.
I can NOT allocate all buffers as maximal size at beginning, because that will exceed the total memory of system too far.
I have to periodically check and re-allocate them respectively.
The problem is that I have to make the client process re-mmap the block, after the server process enlarge (re-allocate) the buffer on the same name.
Regarding performance, is there a way, I can allocate a buffer with some manner likes "lazy allocating"? i.e. it will occupy very small real memory at the beginning but always seat a same virtual address even through I continuously write data resulting that it occupys more and more real memory, and the client does NOT need to re-mmap the buffer, and always can access the data with a fixed virtual address in its own virtual space.
If any, I don't need to do a lot of IPC/locking/sync things.
If any, should I set a very huge swap space?
Related
I've read the docs on the subject.
It says:
When using Buffer.allocUnsafe() to allocate new Buffer instances,
allocations under 4KB are sliced from a single pre-allocated Buffer.
This allows applications to avoid the garbage collection overhead of
creating many individually allocated Buffer instances. This approach
improves both performance and memory usage by eliminating the need to
track and clean up as many individual ArrayBuffer objects.
However, in the case where a developer may need to retain a small
chunk of memory from a pool for an indeterminate amount of time, it
may be appropriate to create an un-pooled Buffer instance using
Buffer.allocUnsafeSlow() and then copying out the relevant bits.
Also I've found this explanation:
Buffer.allocUnsafeSlow is different from Buffer.allocUnsafe() method. In
allocUnsafe() method, if buffer size is less than 4KB than it
automatically cut out the required buffer from a pre-allocated buffer
i.e. it does not initialize a new buffer. It saves memory by not
allocating many small Buffer instances. But if developer need to hold
on some amount of overhead memory for intermediate amount of time,
than allocUnsafeSlow() method can be used.
Though it is hard to understand for me. May you explain it more eloquent and detailed with some examples for both, please?
If a process initially has a number of pages allocated to it in the heap, but a lot of the data in the pages has been deallocated, is there some sort of optimization that the OS does to consolidate the data into one page so that the other pages can be freed?
In general, nothing happens, the heap will continue to have "holes" in it.
Since the (virtual) memory addresses known by a process must remain valid, the operating system cannot perform "heap compaction" on its own. However, some runtimes like .Net do it.
If you are using C or C++, all you can hope for by default is that malloc() will be able to reuse previously deallocated chunks. But if your usage pattern is "allocate a lot of small objects then deallocate half of them at random," the memory utilization will probably not decrease much from the peak.
If a process initially has a number of pages allocated to it in the heap
A process will not initially have pages allocates in a heap.
is there some sort of optimization that the OS does to consolidate the data into one page so that the other pages can be freed
The operating system has no knowledge of user heaps. It allocates pages to the process. What that process does with those pages is up to it (i.e., use them for a heap, stack, code, etc.).
A process's heap manager can consolidate freed chunks of memory. When this occurs, it is normally done to fight heap fragmentation. However, I have never seen a heap manager on a paging system that unmaps pages once they are mapped by the operating system.
The heap of a process never has holes on it. The heap is part of the data segment allocated to a process, that grows dynamically upwards to the top of the stack segment, basically with the use of the sbrk(2) system call (that fixes a new size to the data segment) so the heap is a continuous segment (at least in terms of virtual address space) of allocated pages. malloc(3) never returns the heap space (or part of it) to the system. See malloc(3) for info about this. While there are memory allocators that allow a process to have several heaps (by means of allocating new memory segments, by use of the mmap(2) system call) the segments allocated by a memory allocator are commonly never returned back to the system.
What happens is that the memory allocator reuses the heap space allocated with sbrk(2) and mmap(2) and manages memory for being reused, but it is never returned back to the system.
But don't fear, as this is handled in a good and profitable way by the system, anyway.
That should not affect the overall system management, except from the fact that it consumes virtual address space, and probably page contents will end in the swap device if you don't use them until the process references them again and makes the system to reload them from the swap device(s). If your process doesn't reuse the holes it creates in the heap, the most probable destination is for the system to move them to the swap device and continue reusing it for other processes.
At this moment, I don't know if the system optimices swap allocation by not swapping out zeroed pages, as it does, for example, with text segments of executables (they never go to a swap device, because their contents are already swapped off in the executable file ---this was the reason you couldn't erase in ancient unices a program executable, or the reason there's not need anymore to use the sticky bit in frequently used programs---) but I think it doesn't (and the reason is that it's most improbable the unused pages will be zeroed by the application)
Be warned only in the case you have a 15Gb single process' heap use in your system and 90% of heap use is not in use most of the time. But think better in optimising the allocation resources because a process that consumes 15Gb of heap while most of the time 90%+ is unused, seems to be a poor design. If you have no other chance, simply provide enough swap space to your system to afford that.
I have a question regarding the writeback of the dirty pages. If a portion of page data is modified, will the writeback write the whole page to the disk, or only the partial page with modified data?
The memory management hardware on x86 systems has a granularity of 4096 bytes. This means: It is not possible to find out which bytes of a 4096-byte page are really changed and which ones are unchanged.
Theoretically the disk driver system could check if bytes have been changed and not write the 512-byte blocks that have not been changed.
However this would mean that - if the blocks are no longer in disk cache memory - the page must be read from hard disk to check if it has changed before writing.
I do not think that Linux would do this in that way because reading the page from disk would cost too much time.
Upon EACH hardware interrupt, the CPU would like to write as much data as possible that the harddisk controller can handle - this size is defined by us as the blksize (or ONE sector, in Linux):
http://en.wikipedia.org/wiki/Disk_sector
https://superuser.com/questions/121252/how-do-i-find-the-hardware-block-read-size-for-my-hard-drive
But waiting too long for SINGLE interrupt for a large file can make the system appear unresponsive, so it is logical to break the chunks into smaller size (like 512bytes) so that the CPU can handle other tasks while transferring each 512 bytes down. Therefore, whether u changed one byte or 511 bytes, so long as it is within that single block, all data get written at the same time. And throughout linux kernel, flagging the blocks as dirty for write or not, all goes by the single unique identifier: sector number, so anything smaller than sector size is too difficult for efficient management.
All these said, don't forget that the harddisk controller itself also has a minimum block size for write operation.
As process has virtual memory which is copied into RAM during run time. As given in the previous post.
Which part of process virtual memory layout does mmap() uses?
I have following doubles :
If memory mapping is inside unallocated memory and it is inside process's virtual memory. As virtual memory helps to avoid one process to touch other process's virtual memory. Then how can memory mapping is used for Interprocess Communication(IPC)?
In OS like Linux, whether has each individual process separate section of heap, stack and memory mapping or all processes have one common section for heap, stack and MMAP?
Example :
if there are P1,P2 and P3 processes are running on linux OS. will all have common table as given in picture or each individual task have separate table to each section.
In 32 bit system, 2^32=4 gigabytes of virtual memory is possible and 1G byte is reserved for kernel and 3 gigabytes for userspace applications. can each individual process have up to 3 gigabytes of virtual memory or sum of all userspace applications size could be 3 gigabytes (i.e virtual memory size of (P1+P2+P3)<=3 gigabytes)?
--
Learner
Using memory mapping for IPC works by mapping the same range of physical memory into two or more virtual address ranges in different processes. This works for communication because both processes are using the exact same memory cells (although they might "see" them differently, at different addresses). You change a value in one mapping, and it is instantly visible in the other mapping in a different process because it is the very same memory.
Every process has its own independent stack and heap. The OS does not care about that at all, it only cares about pages. The heap and the stack are things that are implemented by the application (via the runtime). When you call a function like malloc, the allocator in the runtime either returns a block that it already had reserved earlier or one that it has recylced (you called free earlier), or it asks the OS to reserve some more memory (sbrk or mmap). When you first access this memory, the OS sees a page fault and verifies that you are allowed to access this location (because you've reserved it) and then provides a valid page.
Every process can use (as in "reserve") the whole available address space (3GiB in your example). This does not interfere with any other process. Note that due to fragmentation and alignment, and because your executable and the stack take away a little bit, you will in practice not be able to allocate the full 3 GiB, but you can get close to it.
All processes together can use as much virtual memory as is available on the system (physical RAM plus swap space), but they can only use as much as there is physical memory available at the same time (minus a little bit for this and that, like unpageable kernel memory and such).
As program is stored on flash/disk. For it execution, program is loaded into virtual memory and is mapped to RAM by virtual manager. During its execution process is in RAM. Then where does virtual memory exist (where it has all .text, .data, .stack, .heap)?
The virtual memory is a view of the RAM plus maybe some swap space provided by a virtual memory manager. Modern OSs have virtual memory managers and provide virtual memory to processes so that the executing program can behave as if it had a contiguous address space whose size is not limited by the actual RAM. The pages or blocks making up the virtual memory can be mapped anywhere in the RAM, so that contiguos virtual pages need to be stored in contiguos RAM areas. Or they can be swapped out to page space or swap space, waiting there until needed, whereupon they're read by the OS and mapped to some RAM page.
When you say
During its execution process is in RAM.
This is not entirely correct. Some or all memory pages that belong to the process may be swapped out, as explained.
One more word concerning the answers and comments that say that "virtual" means it doesn't exist. This makes no sense. On the contrary, according to Webster:
being such in essence or effect ...
Hence virtual memory is something (therefore, it exists!) that behaves as if it were memory.
Virtual memory is just like an illusion of RAM. It uses paging to acquire additional RAM that could be used by the processes in operating system.
Virtual memory means memory you can access with "normal" momory access methods, although it isn't clear where the data is actually stored.
It may be
actually in RAM
in a swap area
in another file (memory mapped file)
and access to it will be handled appropriately.
It is a layer of, well, virtualization so that you as a programmer don't have to worry about where the data is actually put.
The original purpose was mainly to be able to provide more memory to processes than we actually have and to extend it with means of swap space, but there are even more:
The OS is free to use the RAM for whatever it seems necessary, e. g. caching. Under some circumstances, it may be more effective to use RAM for cache than for holding parts of a program which hasn't been used for a long time.
Provide additional memory to a program when it requests it: if you call malloc(), the program's library may request the OS to provide a part of memory which can be attached seamlessly into the address space.
Avoid stack overflow: if the stack grows larger and larger, the respective memory section may be extended as well transparently so that the program won't have to worry about it.
A system can even do "overcommitment" of memory: if a process requests a large amount of memory, the OS may say "yes, ok", i. e. provide the memory to the program. That means in the first place "allow the program to access a certain address space area", but this address space is not immediately backed by memory. Only as soon as the program accesses this memory the mapping will be done, and if this cannot be fulfilled, the program is crashed by the Out of emory killer (at least, under Linux).
All this works by page-wise (1 page = 4 kiB) assignment of physical memory to a program, viewed via the program's address space, and this in the amount and frequency as it is needed.