When I searched about "What happens if malloc and exit with not free?", I could find answers saying "Today, OS will recover all the allocated memory space after a program exit".
In that answer, what is the meaning of "recover"?.
OS just delete it's PCB and page table when process exit doesn't it?
Are there additional tasks OS has to do for complete termination of process?
When a program starts, the OS allocates some memory to it. During the programs execution the program can request more blocks of memory from the OS and it can release them as well when it doesn't need them any more. When the program exits, all the memory used by it is returned to the OS.
The pair malloc()/free() (and their siblings and equivalents) do not interact with the OS1. They manage a block of memory (called "heap") the program already got from the OS when it was launched.
All in all, from the OS point of view it doesn't matter if your program uses free() or not. For the program it's important to use free() when a piece of memory is not needed to let further allocations succeed (by reusing the freed memory blocks).
1 This is not entirely true. The implementation of malloc() may get more memory blocks from the OS to extend the heap when it is full but this process is transparent to the program. For the program's point of view, malloc() and free() operate inside a memory block that already belongs to the program.
The operating system allocates and manages memory pages in a process. As part of the process cleanup at exit, the operating system has to deallocate the pages assigned to a process. This includes the page tables, page file space, and physical page frames mapped to logical pages. This takes is complicated because multiple processes may map to the same physical page frames which requires some form of reference counting.
The heap is just memory. The operating system has no knowledge whatsoever of process heaps. Inside malloc (and similar functions), there will be calls to operating system services to map pages to the process address space. The operating system creates the pages but does not care what the pages are used for.
If you do malloc's without corresponding free's your process will keep requesting more and more pages from the operating system (until you reach the point where the system will fail to allocate more pages). All you are doing is screwing up the heap within your own process.
When the process exits, the operating system will just get rid of the the pages allocated to the heap and your application's failure to call free will cause no problem to the system at all.
Thus, there is a two level system at work. malloc allocates bytes. The operating system allocates pages.
Related
My understanding is that kmalloc() allocates from anonymous memory. Does this actually reserve the physical memory immediately or is that going to happen only when a write page fault happens?
kmalloc() does not usually allocate memory pages(1), it's more complicated than that. kmalloc() is used to request memory for small objects (smaller than the page size), and manages those requests using already existing memory pages, similarly to what libc's malloc() does to manage the heap in userspace programs.
There are different allocators that can be used in the Linux kernel: SLAB, SLOB and SLUB. Those use different approaches for the same goal: managing allocation and deallocation of kernel memory for small objects at runtime. A call to kmalloc() may use any of the three depending on which one was configured into the kernel.
A call to kmalloc() does not usually reserve memory at all, but rather just manages already reserved memory. Therefore memory returned by kmalloc() does not require a subsequent page fault like a page requested through mmap() normally would.
Copy on Write (CoW) is a different concept. Although it's still triggered through page faults, CoW is a mechanism used by the kernel to save space sharing existing mappings until they are modified. It is not what happens when a fault is triggered on a newly allocated memory page. A great example of CoW is what happens when the fork syscall is invoked: the process memory of the child process is not immediately duplicated, instead existing pages are marked as CoW, and the duplication only happens at the first attempt to write.
I believe this should clear any doubt you have. The short answer is that kmalloc() does not usually "reserve the physical memory immediately" because it merely allocates an object from already reserved memory.
(1) Unless the request is very large, in which case it falls back to actually allocating new memory through alloc_pages().
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.
If I have a process, which has been allocated with some space in RAM. If the process creates a thread (it has too, in fact), the thread will also need some space for its execution. Won't it?
So will it increase the size of the space which has been allocated to that process, or space for thread will be created somewhere else? IF Yes, where on RAM, need it to be contigious with the space that has been possessed by the process?
There'll be some overhead in the scheduler (in the kernel) somewhere since it needs to maintain information about the thread.
There'll also be some overhead in the process-specific area as well since you'll need a stack for each thread and you don't want to go putting stuff into the kernel-specific space when the user code needs to get at.
All modern operating systems and for quite some time now, separate between the memory needed by a process and the memory physically allocated on the RAM.
The OS created a large virtual address space for each process. That address space is independent of how many threads are created inside each process.
In Windows for example, and for optimization reasons, part of that address space is reserved for OS and kernel libraries and is shared amongst all processes for efficiency.
The other part is dedicated to the application user code and libraries.
Once a process logistics and resources are created, the process now is ready to start and that will happen through starting the first thread in the process that will start executing the process main entry point.
For a thread to start execute, it needs a stack amongst other requirements. In Windows, the default size of that stack is about 1 MB. It means, if not changed, each thread will require about 1 MB of memory for its own housekeeping. (stack, TLS, etc....)
When the process needs memory to be allocated, the OS decides the how this memory is going to be allocated physically on the RAM. The process/ application does not see in physical RAM addresses. It only sees virtual addresses from the virtual space assigned to each process.
The OS uses a page file located on the disk to assist with memory requests in addition to the RAM. Less RAM means more pressure on the Page file. When the OS tries to find a piece of memory that's not in the RAM, it will try to find in the page file, and in this case they call it a page miss.
This topic is very extensive but tried to give an overview as much as I can.
From my understanding, when a process is under execution it has some amount of memory at it's disposal. As the stack increases in size it builds from one end of the process (disregarding global variables that come before the stack), while the heap builds from another end. If you keep adding to the stack or heap, eventually all the memory will be used up for this process.
How does the amount of memory the process is given get determined? I can only imagine it depends on a bunch of different variables, but an as-general-as-possible response would be great. If things have to get specific, I'm interested in linux processes written in C++.
On most platforms you will encounter, Linux runs with virtual memory enabled. This means that each process has its own virtual address space, the size of which is determined only by the hardware and the way the kernel has configured it.
For example, on the x86 architecture with a "3/1" split configuration, every userspace process has 3GB of address space available to it, within which the heap and stack are allocated. This is regardless of how much physical memory is available in the system. On the x86-64 architecture, 128TB of address space is typically available to each userspace process.
Physical memory is separately allocated to back that virtual memory. The amount of this available to a process depends upon the configuration of the system, but in general it's simply supplied "on-demand" - limited mostly how much physical memory and swap file space exists, and how much is currently in use for other purposes.
The stack does not magically grow. It's size is static and the size is determined at linking time. So when you take enough space from the stack, it overflows (stack overflow ;)
On the other hand, the heap area 'magically' grows. Meaning that when ever more memory is needed for heap, the program asks operating system for more memory.
EDIT: As Mat pointed out below, the stack actually can increase during runtime on modern operating systems.
I am wondering if i new some object but forget to delete it, when the process exit, will the leaked memory be returned to the OS?
This isn't so mush a C++ question as an operating system question.
All of the operating systems that I have knowledge of will reclaim conventional memory that had been allocated. That's because the allocation generally comes from a processes private address space which will be reclaimed on exit.
This may not be true for other resources such as shared memory. There are implementations that will not release shared memory segments unless you specifically mark them for deletion before your process exits (and, even then, they don't get deleted until everyone has detached).
For most modern operating systems (most flavors of unix and anything running in protected memory under x86), the memory allocation occurs within a program's heap (either through malloc for C or new/delete for C++). So when the program exits, then the memory will be released for use elsewhere.