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Thread: Will XP or Vista 32-bit confuse with 8GB installed?

  1. #1
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    Will XP or Vista 32-bit confuse with 8GB installed?

    My next build will have 8GB due to specific apps needing lots of RAM. That app will be running under a 64bit OS.

    However, I will be multibooting and playing games in XP (or Vista) 32bit.

    Now, if I have 4*2GB installed, will XP or Vista 32bit still work properly even if it can't address all the memory?

    I don't mind losing access to half of the memory when having booted into a 32-bit OS for gaming. But it has to be stable (i.e. no funny OS or game issues due to there being more RAM installed than the OS can address).

    Anybody actually tried this? (i.e. no hearsay, but real testing)?

  2. #2
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    There wont be a problem as your O.S will only map the first 4GB found.

  3. #3
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    What games require 32bit OS?

  4. #4
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    All the games that I want to play fast.

    64bit Vista is slow as pokes for games, even when tweaked according to all the benchmarks I've seen. Not yet seen post sp1 tests though, but I doubt it radically changes the picture.

  5. #5
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    just play them on XP64 - imo this is the best OS out there
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  6. #6
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    Quote Originally Posted by Demo View Post
    There wont be a problem as your O.S will only map the first 4GB found.
    wrong

    32bit Os can only map 3 gigs of ram not 4gb.

    the memory hole remapping feature in some bios's claim that it allows the full 4 gigs of ram, but its doesnt work.
    there are reg tweaks out there that claim they can make all 4 show up but they dont work.

    if your system can handle the 8 gigs windows should be ok.
    but your 6 gigs will be invisible. thats 3 sticks of ram.
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  7. #7
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    Quote Originally Posted by Lestat View Post
    wrong

    32bit Os can only map 3 gigs of ram not 4gb.

    the memory hole remapping feature in some bios's claim that it allows the full 4 gigs of ram, but its doesnt work.
    there are reg tweaks out there that claim they can make all 4 show up but they dont work.

    if your system can handle the 8 gigs windows should be ok.
    but your 6 gigs will be invisible. thats 3 sticks of ram.
    What Demo said was technically correct, 32-bit windows will only map the first 4GB. However, the top level is reserved to hardware addressing, so you end up with 3GB - 3.8GB

  8. #8
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    Quote Originally Posted by Lestat View Post
    wrong

    32bit Os can only map 3 gigs of ram not 4gb.
    Wrong.

    If a computer has a 32-bit OS, it will therefore use 32-bit addressing. The 32-bit wide address allows the processor to address 2^32 bytes of memory, which is exactly 4,294,967,296 bytes, or four gigabytes. As said above, the amount seen is often less than this, due to memory mapped I/O. Memory-mapped I/O places memory which is visible to a peripheral on the system bus (like a graphics card or HDD) in the address space of the processor. To communicate with the device, the processor can simply write the data into the range of memory the device has mapped into its address space. While significantly faster than using a specialised bus for I/O access, this means that the amount of DIMM RAM that your system can see is equal to 4GB minus all peripheral memory.
    Last edited by SoulsCollective; 03-22-2008 at 03:57 AM.
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  9. #9
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    Quote Originally Posted by SoulsCollective View Post
    Wrong.

    If a computer has a 32-bit OS, it will therefore use 32-bit addressing. The 32-bit wide address allows the processor to address 2^32 bytes of memory, which is exactly 4,294,967,296 bytes, or four gigabytes. As said above, the amount seen is often less than this, due to memory mapped I/O. Memory-mapped I/O places memory which is visible to a peripheral on the system bus (like a graphics card or HDD) in the address space of the processor. To communicate with the device, the processor can simply write the data into the range of memory the device has mapped into its address space. While significantly faster than using a specialised bus for I/O access, this means that the amount of DIMM RAM that your system can see is equal to 4GB minus all peripheral memory.

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