Why is it necessary that we need to bring program in main memory from secondary memory for execution?
Why cant we execute program from secondary memory?
Though, it may not be possible currently, but is it possible in future, somehow by some mechanism, that we can execute the program from secondary memory directly?
Almost all modern CPUs execute instructions by fetching them from an address in main memory identified by the instruction pointer register, loading the referenced memory through one or more cache levels before the portion of the CPU that executes the instruction even starts its work. Designing a CPU that could, for example, fetch instructions directly from a disk or network stream would be a rather large project, and performance would likely be pathetic. There's a reason you have a main memory that operates orders of magnitude faster than disk/network access, and caches between that and the actual execution cores that are orders of magnitude faster even than the main memory...
Mostly some parts of the program is required to be accessed multiple times during the execution of the program. Reading from secondary memory every single time we needed the particular data would obviously require a lot of time.
It is better to load the program in a faster memory i.e. Main memory , so that whenever a part of program is required it can be accessed much faster. Similarly, more frequently used variables are stored in the cache memory for even faster access. It;s all about speed.
If could somehow develop affordable secondary memories that have speed as fast as the main memory, we could do without copying the whole program into main memory. However, we would still need some memory to store the temporaries during the program execution.
Main memory is used to distinguish it from external mass storage devices such as hard drives.
Another term for main memory is RAM. The computer can manipulate only data that is in main memory.
So, every program you execute and every file you access must be copied from a
storage device into main memory.The amount of main memory on a computer is crucial because it determines
how many programs can be executed at one time and how much data can be readily available to a program.
Related
Windows, starting with a certain unspecified update of Windows 8.1, has the excellent OfferVirtualMemory and ReclaimVirtualMemory system calls which allow memory regions to be "offered" to the OS. This removes them from the working set, reduces the amount of physical memory usage that is attributed to the calling process, and puts them onto the standby memory list of the program, but without ever swapping out the contents anywhere.
(Below is a brief and rough explanation of what those do and how standby lists work, to help people understand what kind of system call I'm looking for, so skip ahead if you already know all of this.)
Quick standby list reference
Pages in the standby list can be returned back to the working set of the process, which is when their contents are swapped out to disk and the physical memory is used for housing a fresh allocation or swapping in memory from disk (if there's no available "dead weight" zeroed memory on the system), or no swapping happens and the physical memory is returned to the same virtual memory region they were first removed from, sidestepping the swapping process while still having reduced the working set of the program to, well, the memory it's actively working on, back when they were removed from the working set and put into the standby list to begin with.
Alternatively, if another program requests physical memory and the system doesn't have zeroed pages (if no program was closed recently, for example, and the rest of RAM has been used up with various system caches), physical memory from the standby list of a program can be zeroed, removed from the standby list, and handed over to the program which requested the memory.
Back to memory offering
Since the offered memory never gets swapped out if, upon being removed from the standby list, it no longer belongs to the same virtual memory segment (removed from standby by anything other than ReclaimVirtualMemory), the reclamation process can fail, reporting that the contents of the memory region are now undefined (uninitialized memory has been fetched from the program's own standby list or from zeroed memory). This means that the program will have to re-generate the contents of the memory region from another data source, or by rerunning some computation.
The practical effect, when used to implement an intelligent computation cache system, is that, firstly, the reported working set of the program is reduced, giving a more accurate picture of how much memory it really needs. Secondly, the cached data, which can be re-generated from another region of memory, can be quickly discarded for another program to use that cache, without waiting for the disk (and putting additional strain on it, which adds up over time and results in increased wear) as it swaps out the contents of the cache, which aren't too expensive to recreate.
One good example of a use case is the render cache of a web browser, where it can just re-render parts of the page upon request, and has little to no use in having those caches taking up the working set and bugging the user which high memory usage. Pages which aren't currently being shown are the moment where this approach may give the biggest theoretical yield.
The question
Do Linux and macOS have a comparable API set that allows memory to be marked as discardable at the memory manager's discretion, with a fallible system call to lock that memory back in, declaring the memory uninitialized if it was indeed discarded?
Linux 4.5 and later has madvise with the MADV_FREE, the memory may be replaced with pages of zeros anytime until they are next written.
To lock the memory back in write to it, then read it to check if it has been zeroed. This needs to be done separately for every page.
Before Linux 4.12 the memory was freed immediately on systems without swap.
You need to take care of compiler memory reordering so use atomic_signal_fence or equivalent in C/C++.
I have stumbled not once into a term "non coherent" and "coherent" memory in the
tech papers related to graphics programming.I have been searching for a simple and clear explanation,but found mostly 'hardcore' papers of this type.I would be glad to receive layman's style answer on what coherent memory actually is on GPU architectures and how it is compared to other (probably not-coherent) memory types.
Memory is memory. But different things can access that memory. The GPU can access memory, the CPU can access memory, maybe other hardware bits, whatever.
A particular thing has "coherent" access to memory if changes made by others to that memory are visible to the reader. Now, you might think this is foolishness. After all, if the memory has been changed, how could someone possibly be unable to see it?
Simply put, caches.
It turns out that changing memory is expensive. So we do everything possible to avoid changing memory unless we absolutely have to. When you write a single byte from the CPU to a pointer in memory, the CPU doesn't write that byte yet. Or at least, not to memory. It writes it to a local copy of that memory called a "cache."
The reason for this is that, generally speaking, applications do not write (or read) single bytes. They are more likely to write (and read) lots of bytes, in small chunks. So if you're going to perform an expensive operation like a memory load or store, you should load or store a large chunk of memory. So you store all of the changes you're going to make to a chunk of memory in a cache, then make a single write of that cached chunk to actual memory at some point in the future.
But if you have two separate devices that use the same memory, you need some way to be certain that writes one device makes are visible to other devices. Most GPUs can't read the CPU cache. And most CPU languages don't have language-level support to say "hey, that stuff I wrote to memory? I really mean for you to write it to memory now." So you usually need something to ensure visibility of changes.
In Vulkan, memory which is labeled by VK_MEMORY_PROPERTY_HOST_COHERENT_BIT means that, if you read/write that memory (via a mapped pointer, since that's the only way Vulkan lets you directly write to memory), you don't need to use functions vkInvalidateMappedMemoryRanges/vkFlushMappedMemoryRanges to make sure the CPU/GPU can see those changes. The visibility of any changes is guaranteed in both directions. If that flag isn't available on the memory, then you must use the aforementioned functions to ensure the coherency of the specific regions of data you want to access.
With coherent memory, one of two things is going on in terms of hardware. Either CPU access to the memory is not cached in any of the CPU's caches, or the GPU has direct access to the CPU's caches (perhaps due to being on the same die as the CPU(s)). You can usually tell that the latter is happening, because on-die GPU implementations of Vulkan don't bother to offer non-coherent memory options.
If memory is coherent then all threads accessing that memory must agree on the state of the memory at all times, e.g.: if thread 0 reads memory location A and thread 1 reads the same location at the same time, both threads should always read the same value.
But if memory is not coherent then threads A and B might read back different values. Thread 0 could think that location A contains a 1, while thread thinks that that location contains a 2. The different threads would have an incoherent view of the memory.
Coherence is hard to achieve with a high number of cores. Often every core must be aware of memory accesses from all other cores. So if you have 4 cores in a quad core CPU, coherence is not that hard to achieve as every core must be informed about the memory accesses addresses of 3 other cores, but in a GPU with 16 cores, every core must be made aware of the memory accesses by 15 other cores. The cores exchange data about the content of their cache using so called "cache coherence protocols".
This is why GPUs often only support limited forms of coherency. If some memory locations are read only or are only accessed by a single thread, then no coherence is required. If caches are small and coherence is not always required but only at specific instructions of the program, then it is possible to achieve correct behavior of the program using cache flushes before or after specific memory accesses.
If your hardware offers both coherent and non-coherent memory types, then you can expect that non-coherent memory will be faster, but if you try to run parallel algorithms using this memory they will fail in really weird ways.
I know when a program first starts, it has massive page faults in the beginning since the code is not in memory, and thus need to load code from disk.
What happens when a program exits? Does the binary stay in memory? Would subsequent invocations of the program find that the code is already in memory and thus not have page faults (assuming nothing runs in between and pages stuff out to disk)?
It seems like the answer is no from running some experiments on my Linux machine. I ran some program over and over again, and observed the same number of page faults every time. It's a relatively quiet machine so I doubt stuff is getting paged out in between invocations. So, why is that? Why doesn't executable get to stay in memory?
There are two things to consider here:
1) The content of the executable file is likely kept in the OS cache (disk cache). While that data is still in the OS cache, every read for that data will hit the cache and the OS will honor the request without needing to re-read the file from disk
2) When a process exits, the OS unmaps every memory page mapped to a file, frees any memory (in general, releases every resource allocated by the process, including other resources, such as sockets, and so on). Strictly speaking, the physical memory may be zeroed, but not quite required (still, the security level of the OS may require to zero a page that is not used anymore - probably Windows NT, 2K, XP, etc, do that - see this Does Windows clear memory pages?). Another invocation of the same executable will create a brand new process which will map the same file in the memory, but the first access to those pages will still trigger page faults because, in the end, it is a new process, a different memory mapping. So yes, the page faults occur, but they are a lot cheaper for the second instance of the same executable compared to the first.
Of course, this is only about the read-only parts of the executable (the segments/modules containing the code and read-only data).
One may consider another scenario: forking. In this case, every page is marked as copy-on-write. When the first write occurs on each memory page, a hardware exception is triggered and intercepted by the OS memory manager. The OS determines if the page in question is allowed to be written (eg: if it is the stack, heap or any writable page in general) and if so, it allocates memory and copies the original content before allowing the process to modify the page - in order to preserve the original data in the other process. And yes, there is still another case - shared memory, where the exact physical memory is mapped to two or more processes. In this case, the copy-on-write flag is, of course, not set on the memory pages.
Hope this clarifies what is going on with the memory pages.
What I highly suspect is that parts, information blobs are not promptly erased from RAM unless there's a new request for more RAM from actually running code. For that part what probably happens is OS reusing OS dependent bits from RAM, on a next execution e.g. I think this is true for OS initiated resources (and probably not for all resources but some).
Actually most of your questions are highly implementation-dependant. But for most used OS:
What happens when a program exits? Does the binary stay in memory?
Yes, but the memory blocks are marked as unused (and thus could be allocated to other processes).
Would subsequent invocations of the program find that the code is
already in memory and thus not have page faults (assuming nothing runs
in between and pages stuff out to disk)?
No, those blocks are considered empty. Some/all blocks might have been overwritten already.
Why doesn't executable get to stay in memory?
Why would it stay? When a process is finished, all of its allocated resources are freed.
One of the reasons is that one generally wants to clear everything out on a subsequent invocation in case their was a problem in the previous.
Plus, the writeable data must be moved out.
That said, some systems do have mechanisms for keeping executable and static data in memory (possibly not linux). For example, the VMS operating system allows the system manager to install executables and shared libraries so that they remain in memory (paging allowed). The same system can be used to create create writeable shared memory allowing interprocess communication and for modifications to the memory to remain in memory (possibly paged out).
According to this article:
/proc/sys/vm/min_free_kbytes: This controls the amount of memory that is kept free for use by special reserves including “atomic” allocations (those which cannot wait for reclaim)
My question is that what does it mean by "those which cannot wait for reclaim"? In other words, I would like to understand why there's a need to tell the system to always keep a certain minimum amount of memory free and under what circumstances will this memory be used? [It must be used by something; don't see the need otherwise]
My second question: does setting this memory to something higher than 4MB (on my system) leads to better performance? We have a server which occasionally exhibit very poor shell performance (e.g. ls -l takes 10-15 seconds to execute) when certain processes get going and if setting this number to something higher will lead to better shell performance?
(link is dead, looks like it's now here)
That text is referring to atomic allocations, which are requests for memory that must be satisfied without giving up control (i.e. the current thread can not be suspended). This happens most often in interrupt routines, but it applies to all cases where memory is needed while holding an essential lock. These allocations must be immediate, as you can't afford to wait for the swapper to free up memory.
See Linux-MM for a more thorough explanation, but here is the memory allocation process in short:
_alloc_pages first iterates over each memory zone looking for the first one that contains eligible free pages
_alloc_pages then wakes up the kswapd task [..to..] tap into the reserve memory pools maintained for each zone.
If the memory allocation still does not succeed, _alloc pages will either give up [..] In this process _alloc_pages executes a cond_resched() which may cause a sleep, which is why this branch is forbidden to allocations with GFP_ATOMIC.
min_free_kbytes is unlikely to help much with the described "ls -l takes 10-15 seconds to execute"; that is likely caused by general memory pressure and swapping rather than zone exhaustion. The min_free_kbytes setting only needs to allow enough free pages to handle immediate requests. As soon as normal operation is resumed, the swapper process can be run to rebalance the memory zones. The only time I've had to increase min_free_kbytes is after enabling jumbo frames on a network card that didn't support dma scattering.
To expand on your second question a bit, you will have better results tuning vm.swappiness and the dirty ratios mentioned in the linked article. However, be aware that optimizing for "ls -l" performance may cause other processes to become slower. Never optimize for a non-primary usecase.
All linux systems will attempt to make use of all physical memory available to the system, often through the creation of a filesystem buffer cache, which put simply is an I/O buffer to help improve system performance. Technically this memory is not in use, even though it is allocated for caching.
"wait for reclaim", in your question, refers to the process of reclaiming that cache memory that is "not in use" so that it can be allocated to a process. This is supposed to be transparent but in the real world there are many processes that do not wait for this memory to become available. Java is a good example, especially where a large minimum heap size has been set. The process tries to allocate the memory and if it is not instantly available in one large contiguous (atomic?) chunk, the process dies.
Reserving a certain amount of memory with min_free_kbytes allows this memory to be instantly available and reduces the memory pressure when new processes need to start, run and finish while there is a high memory load and a full buffer cache.
4MB does seem rather low because if the buffer cache is full, any process that wants an immediate allocation of more than 4MB will likely fail. The setting is very tunable and system-specific, but if you have a few GB of memory available it can't hurt to bump up the reserve memory to 128MB. I'm not sure what effect it will have on shell interactivity, but likely positive.
This memory is kept free from use by normal processes. As #Arno mentioned, the special processes that can run include interrupt routines, which must be run now (as it's an interrupt), and finish before any other processes can run (atomic). This can include things like swapping out memory to disk when memory is full.
If the memory is filled an interrupt (memory management) process runs to swap some memory into disk so it can free some memory for use by normal processes. But if vm.min_free_kbytes is too small for it to run, then it locks up the system. This is because this interrupt process must run first to free memory so others can run, but then it's stuck because it doesn't have enough reserved memory vm.min_free_kbytes to do its task resulting in a deadlock.
Also see:
https://www.linbit.com/en/kernel-min_free_kbytes/ and
https://askubuntu.com/questions/41778/computer-freezing-on-almost-full-ram-possibly-disk-cache-problem (where the memory management process has so little memory to work with it takes so long to swap little by little that it feels like a freeze.)
I have to run some performance tests, to see how my programs work when the system runs out of RAM and the system starts thrashing. Ideally, I would be able to change the amount of RAM used by the system.
I haved tried to by boot my system (running Ubuntu 10.10) in single user mode with a limited amount of physical memory, but with the parameters I used (max_addr=300M, max_addr=314572800 or mem=300M) the system did not use my swap partition.
Is there a way to limit the amount of RAM used by the total system, while still using swap space?
The point is to measure the total running time of each program as a function of the input size. I am not trying to pinpoint performance problems, I am trying to compare algorithms, which means I need accuracy.
Write a simple c program which
Will allocate large amount of memory.
Keep on accessing allocated memory random to try to keep in main memory (in an infinite loop).
Now run this program (one or few processes) so that you allocate enough memory to cause the thrashing of process you are testing.