As I know, heaps such as malloc is stored in dynamic data. But is it right to say
that malloc function call may allocate the memory space in stack area?
malloc is not a feature of MARS or QtSpim.
There is a system call, #9, that mimics *nix sbrk, to allocate additional address space to the simulated process. It returns to you the next address available past the global data (and past any prior sbrk's). (However, it only allocates, and does not return memory the way a proper sbrk would do given a negative value.) Another name for this area of memory is heap / heap memory.
It does not allocate stack space as that is in substantially higher in the address space. On these simulators, the stack is limited to about 4MB, and the memory returned by sbrk does not reach anywhere near into that 4MB.
Heap memory is useful since a heap allocated memory will survive a function call — a function can return data in the heap but cannot return newly allocated data in the stack, since, by definition, a function that allocates stack space must release that exact same amount when it leaves.
Stack space is allocated simply by decrementing the stack pointer (and released by incrementing the stack pointer). By convention, all functions & subroutines agree to leave existing stack memory alone and allocate new stack memory if they want it, though they also must release it before returning to their caller. When those restrictions do not meet requirements, heap memory is a good choice. Global data is also an option for memory that survives a function call, though that can lead to problems with multithreading (and potentially with recursion).
I am using Delphi 7 Enterprise under Windows 7 64 bit.
My computer had 16 GB of RAM.
I try to use kbmMemTable 7.70.00 Professional Edition (http://news.components4developers.com/products_kbmMemTable.html) .
My table has 150,000 records, but when I try to copy the data from Dataset to the kbmMemTable it only copies 29000 records and I get this error: EOutOfMemory
I saw this message:
https://groups.yahoo.com/neo/groups/memtable/conversations/topics/5769,
but it didn't solve my problem.
An out of memory can happen of various reasons:
Your application uses too much memory in general. A 32 bit application typically runs out of memory when it has allocated 1.4GB using FastMM memory manager. Other memory managers may have worse or better ranges.
Memory fragementation. There may not be enough space in memory for a single large allocation that is requested. kbmMemTable will attempt to allocate roughly 200000 x 4 bytes as one single large allocation. As its own largest single allocation. That shouldnt be a problem.
Too many small allocations leading to the above memory fragmentation. kbmMemTable will allocate from 1 to n blocks of memory per record depending on the setting of the Performance property .
If Performance is set to fast, then 1 block will be allocated (unless blobs fields exists, in which case an additional allocation will be made per not null blob field).
If Performance is balanced or small, then each string field will allocate another block of memory per record.
best regards
Kim/C4D
Please explain it nicely. Don't just write definition. Also explain what it does and how is it different from segmentation.
Fragmentation needs to be considered with memory allocation techniques. Paging is basically not a memory allocation technique, but rather a means of providing virtual address spaces.
Considering the comparison with segmentation, what you're probably asking about is the difference between a memory allocation technique using fixed size blocks (like the pages of paging, assuming 4KB page size here) and a technique using variable size blocks (like the segments used for segmentation).
Now, assume that you directly use the page allocation interface to implement memory management, that is you have two functions for dealing with memory:
alloc_page, which allocates a single page and returns a pointer to the beginning of the newly available address space, and
free_page, which frees a single, allocated page.
Now suppose all of your currently available virtual memory is used, but you need to store 1 additional byte. You call alloc_page and get a 4KB block of memory. You only use 1 byte of that huge block, but also the other 4095 bytes are, from the perspective of the allocator, used. If this happens multiple times eventually all pages will be allocated, so further calls to alloc_page will fail. Even if you just need another additional byte (which could be one of the 4095 that got wasted above) the allocator will tell you that you're out of memory. This is internal fragmentation.
If, on the other hand, you would use variable sized blocks (like in segmentation), then you're vulnerable to external fragmentation: Suppose you manage 6 bytes of memory (F means "free"):
FFFFFF
You first allocate 3 bytes for a, then 1 for b and finally 2 bytes for c:
aaabcc
Now you free both a and c, leaving only b allocated:
FFFbFF
You now have 5 bytes of unused memory, but if you try to allocate a block of 4 bytes (which is less than the available memory) the allocation will fail due to the unfavorable placement of the memory for b. This is external fragmentation.
Now, if you extend your page allocator to be able to allocate multiple pages and add alloc_multiple_pages, you have to deal with both internal and external fragmentation.
There is no external fragmentation in paging but internal fragmentation exists.
First, we need to understand what is external fragmentation. External fragmentation occurs when we have a memory to accommodate a process but it's not continuous.
How does it not occur in paging?
Paging divides virtual memory or all processes into equal-sized pages and physical memory into fixed size frames. So you are typically fixing equal size blocks called pages into equal block shaped spaces called frames! Try to visualize and conclude that there can never be external fragmentation.
In the case of segmentation, we divide virtual addresses into different sized blocks that is why there may be the case some blocks in main memory must stick together or compact to make space for the new process! I hope it helps!
When a process is divided into fix sized pages, there is generally some leftover space in the last page(internal fragmentation). When there are many processes, each of their last page's unused area could add up to be greater than or equal to size of one page. Now even if you have to total free size of one page or more but you cannot load a new page because a page has to be continuous. External fragmentation has happened. So, I don't think external fragmentation is completely zero in paging.
EDIT: It is all about how External Fragmentation is defined. The collection of internal fragmentation do not contribute to external fragmentation. External fragmentation is contributed by the empty space which is EXTERNAL to partition(or page). So if suppose there are only two frames in main memory ,say of size 16B, each occupied by only 1B data. The internal fragmentation in each frame is 15B. The total unused space is 30B. Now if you want to load one new page of some process, you will see that you do not have any frame available. You are unable to load a new page eventhough you have 30B unused space. Will you call this as external fragmentation? Answer is no. Because these 15B unused space are INTERNAL to the pages. So in paging, internal fragmentation is possible but not external fragmentation.
Paging allows a process to be allocated physical memory in non-contiguous fashion. I will answer that why external fragmentation can't occur in paging.
External frag occurs when a process, which was allocated contiguous memory , is unloaded from physical memory, which creates a hole (free space ) in the memory.
Now if a new process comes, which requires more memory than this hole, then we won't be able to allocate contiguous memory to that process due to non contiguous nature of free memory, this is called external fragmentation.
Now, the problem above originated due to the constraint of allocating contiguous memory to the process. This is what paging solved by allowing process to get non contiguous physical memory.
In paging, the probability of having external fragmentation is very low although internal fragmentation may occur.
In paging scheme, the whole main memory and the virtual memory is divided into some fixed size slots which are called pages (in case of virtual memory) and page frames (in case of main memory or RAM or physical memory). So, whenever a process is executed in main memory, it occupies the entire space of a page frame. Let us say, the main memory has 4096 page frames with each page frame having a size of 4096 bytes. Suppose, there is a process P1 which requires 3000 bytes of space for its execution in main memory. So, in order to execute P1, it is brought from virtual memory to main memory and placed in a page frame (F1) but P1 requires only 3000 bytes of space for its execution and as a result of which (4096 - 3000 = 1096 bytes) of space in the page frame F1 is wasted. In other words, this denotes the case of internal fragmentation in the page frame F1.
Again, external fragmentation may occur if some space of the main memory could not be included in a page frame. But this case is very rare as usually the size of a main memory, the size of a page frame as well as the total no. of page frames in main memory can be expressed in terms of power of 2.
As far as I've understood, I would answer your question like so:
Why is there internal fragmentation with paging?
Because a page has fixed size, but processes may request more or less space. Say a page is 32 units, and a process requests 20 units. Then when a page is given to the requesting process, that page is no longer useable despite having 12 units of free "internal" space.
Why is there no external fragmentation with paging?
Because in paging, a process is allowed to be allocated spaces that are non-contiguous in the physical memory. Meanwhile, the logical representation of those blocks will be contiguous in the virtual memory. This is what I mean:
A process requires 128 units of space. This is 4 pages as in the previous example. Unregardless of the actual page numbers (formally frame numbers) in the physical memory, you give those pages the numbers 0, 1, 2, and 3. This is the virtual representation that is the defining characteristic of paging itself. Those pages may be 21, 213, 23, 234 in the actual physical memory. But they can really be anything, contiguous or non-contiguous. Therefore, even if paging leaves small free spaces in between used spaces, those small free spaces can still be used together as if they were one contiguous block of space. That's why external fragmentation won't happen.
Frames are allocated as units. If the memory requirements of a process do not happen to coincide with page boundaries, the last frame allocated may not be completely full.
For example, if the page size is 2,048 bytes, a process of 72,766 bytes will need 35 pages plus 1,086 bytes. It will be allocated 36 frames, resulting in internal fragmentation of 2,048 - 1,086 = 962 bytes. In the worst case, a process would need 11 pages plus 1 byte. It would be allocated 11 + 1 frames, resulting in internal fragmentation of almost an entire frame.
Prequisites
POSIX.1 2008 specifies the setrlimit() and getrlimit() functions. Various constants are provided for the resource argument, some of which are reproduced below for easier understaning of my question.
The following resources are defined:
(...)
RLIMIT_DATA
This is the maximum size of a data segment of the process, in bytes. If this limit is exceeded, the malloc() function shall fail with errno set to [ENOMEM].
(...)
RLIMIT_STACK
This is the maximum size of the initial thread's stack, in bytes. The implementation does not automatically grow the stack beyond this limit. If this limit is exceeded, SIGSEGV shall be generated for the thread. If the thread is blocking SIGSEGV, or the process is ignoring or catching SIGSEGV and has not made arrangements to use an alternate stack, the disposition of SIGSEGV shall be set to SIG_DFL before it is generated.
RLIMIT_AS
This is the maximum size of total available memory of the process, in bytes. If this limit is exceeded, the malloc() and mmap() functions shall fail with errno set to [ENOMEM]. In addition, the automatic stack growth fails with the effects outlined above.
Furthermore, POSIX.1 2008 defines data segment like this:
3.125 Data Segment
Memory associated with a process, that can contain dynamically allocated data.
I understand that the RLMIT_DATA resource was traditionally used to denote the maximum amount of memory that can be assigned to a process with the brk() function. Recent editions of POSIX.1 do no longer specify this function and many operating systems (e.g. Mac OS X) do not support this function as a system call. Instead it is emulated with a variant of mmap() which is not part of POSIX.1 2008.
Questions
I am a little bit confused about the semantic and use of the RLIMIT_DATA resource. Here are the concrete questions I have:
Can the stack be part of the data segment according to this specification?
The standard says about RLIMIT_DATA: “If this limit is exceeded, the malloc() function shall fail with errno set to [ENOMEM].” Does this mean that memory allocated with malloc() must be part of the data segment?
On Linux, memory allocated with mmap() does not count towards the data segment. Only memory allocated with brk() or sbrk() is part of the data segment. Recent versions of the glibc use a malloc() implementation that allocates all its memory with mmap(). The value of RLIMIT_DATA thus has no effect on the amount of memory you can allocate with this implementation of malloc().
Is this a violation of POSIX.1 2008?
Do other platforms exhibit similar behavior?
The standard says about RLIMIT_AS: "If this limit is exceeded, the malloc() and mmap() functions shall fail with errno set to [ENOMEM]." As the failure of mmap() is not specified for RLIMIT_DATA, I conclude that memory obtained from mmap() does not count towards the data segment.
Is this assumption true? Does this only apply to non-POSIX variants of mmap()?
FreeBSD also shares the problem of malloc(3) being implemented using mmap(2) in the default malloc implementation. I ran into this when porting a product from FreeBSD 6 to 7, where the switch happened. We switched the default limit for each process from RLIMIT_DATA=512M to RLIMIT_VMEM=512M, i.e. limit the virtual memory allocation to 512MB.
As for whether this violates POSIX, I don't know. My gut feeling is that lots of things violate POSIX and a 100% POSIX compliant system is as rare as a strictly-confirming C compiler.
EDIT: heh, and now I see that FreeBSD's name RLIMIT_VMEM is non-standard; they define RLIMIT_AS as RLIMIT_VMEM for POSIX compatibility.
I checked the FastMM4 with D6.
When I debug a simple application with uses "Forms", I everytime got 3 lines for memory leak.
This application has leaked memory.
The small block leaks are (excluding
expected leaks registered by pointer):
13 - 20 bytes: TObjectList x 3,
Unknown x 3 29 - 36 bytes:
TWinHelpViewer x 1 37 - 52 bytes:
THelpManager x 1
Is this normal?
Which thing causes this?
Thanks:
dd
The RTL/VCL that ships with Delphi 6 contains some memory leaks. In later releases of Delphi the use of FastMM led to these memory leaks being removed from the RTL/VCL.
What you need to do is register these known and expected memory leaks with FastMM. Once you have registered the leaks that FastMM won't report them. Although these leaks are real, they are best ignored for various reasons:
The leaked memory from these known VCL leaks is tiny and doesn't grow during the lifetime of the process.
The memory returned to the system as soon as the process terminates anyway.
Since the leaks are in code beyond your control, there's not a huge amount you can do. You could fix them and use your own version of the VCL units in question, but is it worth it?
The only time these leaks could matter is if you had a DLL which was loaded and unloaded from the same process thousands of times during the lifetime of that process. I don't believe this is a very realistic scenario.
If you don't register the leaks then the FastMM leak reporting becomes largely ineffective because it shows every time. If it shows every time you learn to ignore it. This leak reporting is very valuable, but it is only valuable if it shows leaks that you have some control over.
In my Delphi 6 project I have the following code in my .dpr file:
// Register expected VCL memory leaks caused by Delphi unit HelpIntfs.
FastMM4.RegisterExpectedMemoryLeak(36, 2); // THelpManager x 1, THTMLHelpViewer x 1
FastMM4.RegisterExpectedMemoryLeak(20, 7); // TObjectList x 3, THelpSelector x 1, Unknown x 3
FastMM4.RegisterExpectedMemoryLeak(52); // TWinHelpViewer x 1
I also have the following in a TForm descendant from which all forms in my app descend:
var
ExpectedHelpStringMemoryLeakRegistered: Boolean;
procedure TMyForm.WMHelp(var Message: TWMHelp);
begin
if not (biHelp in BorderIcons) and not ExpectedHelpStringMemoryLeakRegistered then begin
// Register expected VCL memory leaks caused by Delphi unit HelpIntfs.
FastMM4.RegisterExpectedMemoryLeak(44); // TString x 1
ExpectedHelpStringMemoryLeakRegistered := True;
end;
inherited;
end;
Depending on exactly which units you use in the RTL/VCL and how you use them, you may need to register different memory leaks.
I would guess it is normal unless you patched the sources. IIRC, when there was 'memproof', its author 'Atanas Stoyanov' kept a list of bugs that caused memory leaks. The leak in the 'classes.pas', f.i., effected every VCL forms application. Though the product does not exist anymore, you can find the list at 'Automated QA's site. Here's the list for D6.