Will Vista recognize 4GB?

JLGatsby

Weaksauce
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Sep 6, 2005
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I know XP did not properly recognize the last 1GB in a 4GB set up, will Vista be able to recognize 4GB completely?
 
what motherboard do you have? some motherboards do not go over 3GB, if I recall...some could do 4GB.
 
unless i am mistaken... 32bit addressing only allows for 3GB of memory. 64bit allows for a whole lot more than we will have in the forseeable future (40GB or something, i cannot remember). so, if you have an x86-64 or full 64bit processor the answer is: yes, if you install 64bit vista. otherwise... no.
 
byne said:
unless i am mistaken... 32bit addressing only allows for 3GB of memory.

2^32 is 4,294,967,296, which is four gigabytes. 3GB is 3,221,225,472.

LATER: Oh; are you thinking of virtual address space?
 
Edit notice --- I just glanced at Mike's address space and I'm sure it covered everything I mentioned below. I should have read all the things before responding. Anyways. I'd delete this post if I knew how to do it... So I guess just ignore what I said. haha.

Original post ---

32 bit allows 4 GB.

However, Windows XP (32 bit) reserves space for system and virtual addressing. Typically, for example, each program has up to 2 gigabytes of available storage/ram. This is logical ram, not physical ram. The kernel then records what physical ram is used by which process, and maps the programs logical ram into the physical ram. With the case of virtual memory, this logical ram can also be mapped to the hard disk and swapped to physical ram when needed. All of that is hidden from the programmers and is usually handled automatically.

Windows reserves 2 gigabytes for its own use. So to make a long story short, on 32 bit applications, any program can use up to 2 gigs of ram, and the system will reserve 2 gigs of ram... So I do believe if you have 4 gigs of ram it will be used, however, your programs are still limited to 2 gigs of ram due to operating system restrictions. However, you will never be wasting your 4 gigs of ram in a 32 bit world, its just that these kernal/OS operations will be taking/using 2 gigs of that available ram internally.

I don't know how well I explained that but... My guess Vista will be written with 32 bit and 64 bit like XP is done today, and will probably have similar restrictions as the kernal was probably not rewritten from scratch.
 
mikeblas said:
2^32 is 4,294,967,296, which is four gigabytes. 3GB is 3,221,225,472.

LATER: Oh; are you thinking of virtual address space?
that link definatly explained it all if you read down further.
 
byne said:
that link definatly explained it all if you read down further.
Which link? The one I posted to the article I wrote? Thank you!
 
Scoobydo said:
Windows reserves 2 gigabytes for its own use. So to make a long story short, on 32 bit applications, any program can use up to 2 gigs of ram, and the system will reserve 2 gigs of ram... So I do believe if you have 4 gigs of ram it will be used, however, your programs are still limited to 2 gigs of ram due to operating system restrictions. However, you will never be wasting your 4 gigs of ram in a 32 bit world, its just that these kernal/OS operations will be taking/using 2 gigs of that available ram internally.

I don't know how well I explained that but... My guess Vista will be written with 32 bit and 64 bit like XP is done today, and will probably have similar restrictions as the kernal was probably not rewritten from scratch.

Either 2 GB for apps and 2 GB for the OS. Or, 3 GB for apps and 1 GB for the OS, if you've booted with /3GB. Vista won't change this.

Not to make it a semantic argument, but I'd say that the OS can always see 4GB of physical memory. It has to see its own memory and the memory of the processes it is running. The processes? Indeed, they might see less physical address space.
 
mikeblas said:
Either 2 GB for apps and 2 GB for the OS. Or, 3 GB for apps and 1 GB for the OS, if you've booted with /3GB. Vista won't change this.

Not to make it a semantic argument, but I'd say that the OS can always see 4GB of physical memory. It has to see its own memory and the memory of the processes it is running. The processes? Indeed, they might see less physical address space.

if i am understanding what the OP is saying, its not that he cannot fill up this space, but xp does not see it. as in... on the computer properties it says he has 3GB physical. This number does not exclude the memory reserved for the os. the x86-64 duality would change these things, even if he only sees 3GB, on a 64bit os it would open up (again, only if x86-64) as there is nothing in AMD's Clawhammer motherboard specification that would call for any such memory limitation in 64bit mode at such a low level.
so, for clarity, lets go through what it might be:

limitation of the processor: x86 (as in, not Athlon64s)
limitation of the motherboard: older, and some new motherboards had caps to the to increase performance. original athlon boards only allowed for 2GB, then 3GB within the athlon XP generation. these increases did not always cooincide with the release of new processor lines (intel or amd, i just remember approx amd).
limitation of the OS: even with all the hardware to support it, if the os doesn't then the athlon64s default to legacy mode and have old restrictions. again, the os is not the only thing restricting the adressable space.
hardware problem: a bad stick, not seated, bad slot
 
byne said:
if i am understanding what the OP is saying, its not that he cannot fill up this space, but xp does not see it. as in... on the computer properties it says he has 3GB physical. This number does not exclude the memory reserved for the os.

No memory is reserved for the OS. The OS loads, and gives itself what it needs.

byne said:
so, for clarity, lets go through what it might be:

"It" meaning: what might becausing the OP to see 3GB available on a machine with 4GB physical? Your list is correct except for the part about the OS. Windows XP can, given capable hardware and drivers, get really (really!) close to the 4 GB available pyhsical mark, even on Win32.
 
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