No memory although more than 1GB free

In microsoft.public.vc.stl Nick Savoiu
wrote:
"Klaus Bonadt" wrote in message
...
I have the AMD 64 processor, but I need a special 64-bit XP system,
which is not yet available for AMD, correct?

You can download a preview version for free from the MS site.

But anyway, as I have mentioned above there must be at least 2 - 1,8
= 0,2GB unfragmented memory available, otherwise my test program
could not allocate 1,2GB until the whole virtual memory is
occupied.

Each process has a separate virtual address space so fragmentation in
one would not affect another.

It might be a bit off-topic by now, but what is the largest chunk of
'flat memory' I can expect to get on a 32bit Intel system stuffed with
e.g. 3GB of RAM and running not much else than XP?

Andre'

"Klaus Bonadt" wrote in message
...
I have the AMD 64 processor, but I need a special 64-bit XP system, which
is
not yet available for AMD, correct?

You can download a preview version for free from the MS site.

But anyway, as I have mentioned above there must be at least 2 - 1,8 =
0,2GB
unfragmented memory available, otherwise my test program could not
allocate
1,2GB until the whole virtual memory is occupied.

Each process has a separate virtual address space so fragmentation in one
would not affect another.

It seems that for every HeapAlloc there are 8 Bytes additional costs in
terms of memory, but the number of allocations
seem to be not limited.

Of course. The OS needs to do some bookkeeping for each allocation.

Nick

"Klaus Bonadt" wrote in message
...
Thus, each process is able to allocate 2GB at maximum. Indeed, I can
almost
allocate 2GB with my test program, allocating chunks of 1MB.
However, when my crucial application runs out of memory, Task Manager
tells
that there is only 1,6GB in use (see my first mail).

That's why I think there's fragmentation at work. Imagine the
pathological scenario: your whole memory is occupied by 1K allocated
chunk followed by 1023K free chunk, and so on. 1024 such pairs will eat
up 1GB of RAM in a way that only 1MB is actually used, but you cannot
allocate another 1MB chunk. It's an artificial example of course, it's
highly unlikely to occur in practice, but it demonstrates the idea of
fragmentation nicely.

Furthermore, starting
my test program at this point in time, this program is able to
allocate
further 1,2 GB until my whole virtual memory is allocated.

Your test program is a separate process that has its own address space.
Remember - you are not running out of physical RAM, you are running out
of addresses to map it to. Each process has its very own 4GB address
space, independent of any other process. Physical RAM gets mapped to a
process' address space on as-needed basis.

By the way, with Windows virtual memory management, you cannot fragment
physical memory at all, you can only fragment virtual address space. The
way it works, all physical memory is broken into pages 4KB (on some
systems 8KB) large. So is virtual memory. When a region of virual memory
is allocated, each page of virtual memory is backed by a page of
physical memory. Now, virtual addresses within this region need to be
consequtive (because your program expects the address arithmetic to
work), but physical RAM pages don't need to be. The system just picks
free RAM pages lying around and maps them to virtual pages. Moreover, if
the system runs out of RAM, it picks a physical page (mapped to some
virtual page in some process A), saves its contents to disk and reuses
it for another virtual page possibly in a different process B. If
process A later needs to refer to that virtual page, some other physical
page may be assigned to it and contents read back from disk.

Bottom line: there is no permanent attachment of virtual addresses to
physical RAM pages, and there is no requirement that a contiguous
stretch of virtual addresses be backed by a contiguous stretch of
physical addresses. Thus RAM cannot become fragmented, but virtual
address space can.

I have the AMD 64 processor, but I need a special 64-bit XP system,
which is
not yet available for AMD, correct?

Sorry, I don't know.

Even if you have enough address space, it may be fragmented. That
is,
there are many small stretches of unused address space, but none
large
enough to accomodate your allocation request.

What are PVIEW and Task Manager showing, the sum of heap sizes
(including
unused space due to fragmentation) or the sum of actual allocated
memory
(HeapAlloc) which was not freed afterwards?

I believe they show a working set, which is memory that is a) actually
allocated and b) actually loaded in RAM, rather than being swapped to
hard drive. Well, in your case of huge amounts of RAM available, you can
assume that a working set is very close to actual RAM consumption. They
don't show the unused space, because it is, well, unused.

I guess the first case, which means the sum of memory which is
reserved for
the process even the process is not able to allocate due to
fragmentation?

If this were the case, they would always show 2GB - that's the amount of
address space given to every process. Such a display would be pretty
useless, don't you think?
--
With best wishes,
Igor Tandetnik

"For every complex problem, there is a solution that is simple, neat,
and wrong." H.L. Mencken

In microsoft.public.vc.stl Nick Savoiu
wrote:
"Klaus Bonadt" wrote in message
...
I have the AMD 64 processor, but I need a special 64-bit XP system,
which is not yet available for AMD, correct?

You can download a preview version for free from the MS site.

But anyway, as I have mentioned above there must be at least 2 - 1,8
= 0,2GB unfragmented memory available, otherwise my test program
could not allocate 1,2GB until the whole virtual memory is
occupied.

Each process has a separate virtual address space so fragmentation in
one would not affect another.

It might be a bit off-topic by now, but what is the largest chunk of
'flat memory' I can expect to get on a 32bit Intel system stuffed with
e.g. 3GB of RAM and running not much else than XP?

Andre'

"Klaus Bonadt" wrote in message
...
Thus, each process is able to allocate 2GB at maximum. Indeed, I can
almost
allocate 2GB with my test program, allocating chunks of 1MB.
However, when my crucial application runs out of memory, Task Manager
tells
that there is only 1,6GB in use (see my first mail).

That's why I think there's fragmentation at work. Imagine the
pathological scenario: your whole memory is occupied by 1K allocated
chunk followed by 1023K free chunk, and so on. 1024 such pairs will eat
up 1GB of RAM in a way that only 1MB is actually used, but you cannot
allocate another 1MB chunk. It's an artificial example of course, it's
highly unlikely to occur in practice, but it demonstrates the idea of
fragmentation nicely.

Furthermore, starting
my test program at this point in time, this program is able to
allocate
further 1,2 GB until my whole virtual memory is allocated.

Your test program is a separate process that has its own address space.
Remember - you are not running out of physical RAM, you are running out
of addresses to map it to. Each process has its very own 4GB address
space, independent of any other process. Physical RAM gets mapped to a
process' address space on as-needed basis.

By the way, with Windows virtual memory management, you cannot fragment
physical memory at all, you can only fragment virtual address space. The
way it works, all physical memory is broken into pages 4KB (on some
systems 8KB) large. So is virtual memory. When a region of virual memory
is allocated, each page of virtual memory is backed by a page of
physical memory. Now, virtual addresses within this region need to be
consequtive (because your program expects the address arithmetic to
work), but physical RAM pages don't need to be. The system just picks
free RAM pages lying around and maps them to virtual pages. Moreover, if
the system runs out of RAM, it picks a physical page (mapped to some
virtual page in some process A), saves its contents to disk and reuses
it for another virtual page possibly in a different process B. If
process A later needs to refer to that virtual page, some other physical
page may be assigned to it and contents read back from disk.

Bottom line: there is no permanent attachment of virtual addresses to
physical RAM pages, and there is no requirement that a contiguous
stretch of virtual addresses be backed by a contiguous stretch of
physical addresses. Thus RAM cannot become fragmented, but virtual
address space can.

I have the AMD 64 processor, but I need a special 64-bit XP system,
which is
not yet available for AMD, correct?

Sorry, I don't know.

Even if you have enough address space, it may be fragmented. That
is,
there are many small stretches of unused address space, but none
large
enough to accomodate your allocation request.

What are PVIEW and Task Manager showing, the sum of heap sizes
(including
unused space due to fragmentation) or the sum of actual allocated
memory
(HeapAlloc) which was not freed afterwards?

I believe they show a working set, which is memory that is a) actually
allocated and b) actually loaded in RAM, rather than being swapped to
hard drive. Well, in your case of huge amounts of RAM available, you can
assume that a working set is very close to actual RAM consumption. They
don't show the unused space, because it is, well, unused.

I guess the first case, which means the sum of memory which is
reserved for
the process even the process is not able to allocate due to
fragmentation?

If this were the case, they would always show 2GB - that's the amount of
address space given to every process. Such a display would be pretty
useless, don't you think?
--
With best wishes,
Igor Tandetnik

"For every complex problem, there is a solution that is simple, neat,
and wrong." H.L. Mencken

Xref: kermit microsoft.public.vc.language:144401 microsoft.public.vc.stl:18865 microsoft.public.windowsxp.perform_maintain:155619

But anyway, as I have mentioned above there must be at least 2 - 1,8 =
0,2GB
unfragmented memory available, otherwise my test program could not
allocate
1,2GB until the whole virtual memory is occupied.

Each process has a separate virtual address space so fragmentation in one
would not affect another.

In my current situation, my application runs out of memory due to a memory
claim of 0.5MB RAM.
Let us assume this is due to fragmented memory, i.e. there is no contiguous
range with more than 0.5MB RAM in the lower 2 GB of virtual address space.
The application stops with popping a message box.
Now I start another application, which allocates in a loop chunks of memory,
each chunk is 1MB. The application does this up to 1.2 GB RAM. I have 3 GB
RAM in total (fixed 1 GB page size). Thus, there was 3 - 1.2 = 1.8GB for my
crucial application plus operating system.
Even the operating system does not need any memory, there are at least 2GB -
1.8GB = 0.2GB in the lower address space available with memory chunks
greater or equal 1MB (otherwise the test program would not be able to
allocate in this area 1MB chunks)!

Excuse me for my questions, but I still do not understand what is going on.

It seems that for every HeapAlloc there are 8 Bytes additional costs in
terms of memory, but the number of allocations
seem to be not limited.

Of course. The OS needs to do some bookkeeping for each allocation.

OK, but 133,164,202 allocations is probably more than enough, it is
sufficient for allocating the maximum number of chunks with 2Byte size.
Only when freeing memory does not result in a free'd entry in the book, this
could be a limited resource. I could try to simulate memory fragmentation
with my test program.

Thanks for your patience with me!
Regards,
Klaus

Xref: kermit microsoft.public.vc.language:144401 microsoft.public.vc.stl:18865 microsoft.public.windowsxp.perform_maintain:155619

But anyway, as I have mentioned above there must be at least 2 - 1,8 =
0,2GB
unfragmented memory available, otherwise my test program could not
allocate
1,2GB until the whole virtual memory is occupied.

Each process has a separate virtual address space so fragmentation in one
would not affect another.

In my current situation, my application runs out of memory due to a memory
claim of 0.5MB RAM.
Let us assume this is due to fragmented memory, i.e. there is no contiguous
range with more than 0.5MB RAM in the lower 2 GB of virtual address space.
The application stops with popping a message box.
Now I start another application, which allocates in a loop chunks of memory,
each chunk is 1MB. The application does this up to 1.2 GB RAM. I have 3 GB
RAM in total (fixed 1 GB page size). Thus, there was 3 - 1.2 = 1.8GB for my
crucial application plus operating system.
Even the operating system does not need any memory, there are at least 2GB -
1.8GB = 0.2GB in the lower address space available with memory chunks
greater or equal 1MB (otherwise the test program would not be able to
allocate in this area 1MB chunks)!

Excuse me for my questions, but I still do not understand what is going on.

It seems that for every HeapAlloc there are 8 Bytes additional costs in
terms of memory, but the number of allocations
seem to be not limited.

Of course. The OS needs to do some bookkeeping for each allocation.

OK, but 133,164,202 allocations is probably more than enough, it is
sufficient for allocating the maximum number of chunks with 2Byte size.
Only when freeing memory does not result in a free'd entry in the book, this
could be a limited resource. I could try to simulate memory fragmentation
with my test program.

Thanks for your patience with me!
Regards,
Klaus

One other tool useful for a meaningful dump of the process address space
is the '!address' command in cdb/ntsd/windbg.
If will also try to give the usage of the allocated regions, their size,
flags, etc, etc.

--
This posting is provided "AS IS" with no warranties, and confers no rights.
Use of any included script samples are subject to the terms specified at
http://www.microsoft.com/info/cpyright.htm


"Klaus Bonadt" wrote in message
...
It seems that you could have that case. You could try to walk the
virtual
address space and see what's going on. Check out VirtualQueryEx().

DWORD VirtualQuery(
LPCVOID lpAddress, // address of region
PMEMORY_BASIC_INFORMATION lpBuffer, // information buffer
SIZE_T dwLength // size of buffer
);

Two questions:
1.
How to set lpAddress initially?

2.
I will collect sets of memory regions with different attributes. However,
I
do not know how to interpret these sets.
Furthermore, I guess I will end up with the same amount of memory, which
"Process viewer" is presenting.

Regards,
Klaus

Sounds like you are using a old application. What is it you are trying to
run..?

Self-written data-warehouse application!

Regards,
Klaus

Sounds like you are using a old application. What is it you are trying to
run..?

Self-written data-warehouse application!

Regards,
Klaus

"Klaus Bonadt" wrote in message
...
Each process has a separate virtual address space so fragmentation in
one
would not affect another.

In my current situation, my application runs out of memory due to a memory
claim of 0.5MB RAM.
[...]
Even the operating system does not need any memory, there are at least
2GB -
1.8GB = 0.2GB in the lower address space available with memory chunks
greater or equal 1MB (otherwise the test program would not be able to
allocate in this area 1MB chunks)!
Excuse me for my questions, but I still do not understand what is going
on.

Not a problem. We're all here to hopefully learn something. Try re-reading
Igor's posting again. He explains quite well what goes on with virtual/real
memory and address spaces. If that doesn't clear things up I'd suggest
picking up an OS design book and reading up on virtual memory and such.

I could try to simulate memory fragmentation with my test program.

This might be not be easy to do as you do not directly control the virtual
addresses at which you make allocations. So the OS could be playing tricks
behind your back.

Nick

In microsoft.public.vc.stl Igor Tandetnik wrote:
That's why I think there's fragmentation at work. Imagine the
pathological scenario: your whole memory is occupied by 1K allocated
chunk followed by 1023K free chunk, and so on. 1024 such pairs will eat
up 1GB of RAM in a way that only 1MB is actually used, but you cannot
allocate another 1MB chunk. It's an artificial example of course, it's
highly unlikely to occur in practice, but it demonstrates the idea of
fragmentation nicely.

Does that mean the allocator does _not_ try to collect chunks of 'almost
the same size' in one place and different sizes in another one? [I am
talking about virtual adresses here]

Andre'

By the way, with Windows virtual memory management, you cannot fragment
physical memory at all, you can only fragment virtual address space. The
way it works, all physical memory is broken into pages 4KB (on some
systems 8KB) large. So is virtual memory. When a region of virual memory
is allocated, each page of virtual memory is backed by a page of
physical memory. Now, virtual addresses within this region need to be
consequtive (because your program expects the address arithmetic to
work), but physical RAM pages don't need to be. The system just picks
free RAM pages lying around and maps them to virtual pages. Moreover, if
the system runs out of RAM, it picks a physical page (mapped to some
virtual page in some process A), saves its contents to disk and reuses
it for another virtual page possibly in a different process B. If
process A later needs to refer to that virtual page, some other physical
page may be assigned to it and contents read back from disk.

Thanks a lot, Igor, I got it!

"Klaus Bonadt" wrote in message
...
But anyway, as I have mentioned above there must be at least 2 -
1,8 =
0,2GB
unfragmented memory available, otherwise my test program could not
allocate
1,2GB until the whole virtual memory is occupied.

Each process has a separate virtual address space so fragmentation
in one
would not affect another.

In my current situation, my application runs out of memory due to a
memory
claim of 0.5MB RAM.
Let us assume this is due to fragmented memory, i.e. there is no
contiguous
range with more than 0.5MB RAM in the lower 2 GB of virtual address
space.
The application stops with popping a message box.
Now I start another application, which allocates in a loop chunks of
memory,
each chunk is 1MB. The application does this up to 1.2 GB RAM.

Which part of "each process has a separate virtual address space" do you
have difficulty understanding? A success or failure of a memory
allocation in one process tells you nothing about the state of the
virtual address space in another. It's like saying: "I have machine A
where memory allocation fails. Then I run a test program on machine B,
and it can successfully allocate plenty of memory. So enough memory
should be avaiable on machine A, right?"
--
With best wishes,
Igor Tandetnik

"For every complex problem, there is a solution that is simple, neat,
and wrong." H.L. Mencken

"Klaus Bonadt" wrote in message
...
Each process has a separate virtual address space so fragmentation in
one
would not affect another.

In my current situation, my application runs out of memory due to a memory
claim of 0.5MB RAM.
[...]
Even the operating system does not need any memory, there are at least
2GB -
1.8GB = 0.2GB in the lower address space available with memory chunks
greater or equal 1MB (otherwise the test program would not be able to
allocate in this area 1MB chunks)!
Excuse me for my questions, but I still do not understand what is going
on.

Not a problem. We're all here to hopefully learn something. Try re-reading
Igor's posting again. He explains quite well what goes on with virtual/real
memory and address spaces. If that doesn't clear things up I'd suggest
picking up an OS design book and reading up on virtual memory and such.

I could try to simulate memory fragmentation with my test program.

This might be not be easy to do as you do not directly control the virtual
addresses at which you make allocations. So the OS could be playing tricks
behind your back.

Nick