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Does winxp count v/c mem as system memory

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Joined
23 Feb 2008
Posts
81
Hello.
My friend has winxp32,and has 4gb mem,and also runs a GTX 280 1gb mem v/c.
He is wondering because winxp 32 will only run 3.odd gb mem,that the 1gb mem on his 280v/c isn't being utilised,is this true?

He runs an E8500 cpu.
Thanks.
 
Yes it will, but will discard the RAM before the g/c memory IIRC.
kae2
So what you are saying,it will use the v/c memory first,before the mobo mem
if that is how i understand it,so windows counts the mobo mem and the v/c mem altogether.
He says the sims he plays doesn't run on full max,,so he is thinking of getting just 2 gb of mem,instead of the 4gb mem,and hoping the system will use it all
including the 1gb of the v/c,and run the sims at max.
Thanks for posting.
 
kae2 is correct.

So say you make the epic noob erro of having 4GB of RAM and then 1GB of VRAM when only 3.2GB is detectable.

1GB of that 3.2GB will be VRAM so in reality you will only have 2.2GB of RAM.

Bad times!
 
Only reason why he is only seeing around 3gb of RAM because of 32 bit XP limitations. You need a 64 bit operating system to use the full 4gb.
 
The GPU driver does not map the entire 1GB of VRAM into the 32-bit address space - it will only use a small window of the 32-bit address space (eg 200MB) and then map in parts of the VRAM which the CPU wants to communicate with. It's a kind of window scheme.

For instance, if the driver wants to load in a texture to the low end of the 1GB of VRAM, it'll map the bottom 200MB into its allocation of the address space. It'll then transfer in the texture. Let's say it then wants to load in a texture to the top of the 1GB of VRAM: it'll unmap the bottom 200MB and then remap the top 200MB to that place, then copy in the texture.
This scheme allows you to use a graphics card with 100GB of VRAM, in a 32-bit address space - it only maps in parts (again, small regions like 200MB) that it actually wants the CPU to access. The GPU can always access all of its internal memory.

The reason you're not getting the full 4GB of RAM is because of the 32-bit OS issue. Other devices want a small share of this 4GB space and they each map their own memory into this space in the same way, slowly eroding the 4GB address space.
 
kae2 is correct.

So say you make the epic noob erro of having 4GB of RAM and then 1GB of VRAM when only 3.2GB is detectable.

1GB of that 3.2GB will be VRAM so in reality you will only have 2.2GB of RAM.

Bad times!

I think he's saying he only has 3G available after 1G has been mapped to the GPU not before.
 
Hello.
My friend has winxp32,and has 4gb mem,and also runs a GTX 280 1gb mem v/c.
He is wondering because winxp 32 will only run 3.odd gb mem,that the 1gb mem on his 280v/c isn't being utilised,is this true?

He runs an E8500 cpu.
Thanks.

32-bit OS can only address 4GB of address space (2^32 = 4GB) ... however this cannot all be RAM since the OS also needs to be able to talk to peripherals (including the gfx card) etc and these have to be mapped into the address space. As a result there's a largish section at the top of the address space that is reserved for these purposes with the result that only the bottom ~3.25GB of the address space maps to RAM - i.e. there's only 3.25GB of addressable RAM. (As an analogy think of phone numbers .... a local number may have 6 digits but you can't have 10^6 phones in that dial code since numbers like 999xxx can't be used, anything starting with 0 is mapped to long distance etc etc).

This 3.25 of addressable RAM is all normal RAM - the RAM on the gfx card is accessed through being mapped through a window in the top "peripheral" region (CPU programs control regs on the gfx card to determine which part of the RAM on the gfx card is seen in this window at any time)

If the gfx were being supplied via integrated graphics in the chipset then in that case part of the RAM would be allocated as gfx RAM which would reduce amount of the 3.25 addressable RAM available to the OS.
 
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