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Special Forums Hardware Filesystems, Disks and Memory Installed memory ≠ usable size? Post 302332683 by CRGreathouse on Thursday 9th of July 2009 10:40:17 PM
Old 07-09-2009
Quote:
Originally Posted by otheus
Right. You were talking to the "Geek Squad" at BestBuy, werent you?
No, why would you say that? Smilie I bought the components for the computer at MicroCenter and assembled it myself. I haven't been to a BestBuy for ages.

I can't tell how you intended the statement since your intonation doesn't come through the text. Are you doubting my statement on 64-bit addressing? My understanding is that AMD64 systems set 16 memory bits (actually I think 16 of the middle bits, rather than just the 16 upper bits) to 1, limiting the actual space that can be accessed to 2^(64-16) = 2^48 = 64 terabytes. Let me try to Google up a source,

OK, I have one:
EmbeddedDeveloper.com - x86 - AMD64 (64-bit)
Actually this gives a limit of 48 bits of virtually-addressed memory and only 40 bits (1 terabyte) of physical memory.

Of course these are all better than Vista's 37 bits (128 GB for Vista Business), though I think that's an artificial restriction.
 

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BITMAP_REMAP(9) 					  Basic Kernel Library Functions					   BITMAP_REMAP(9)

NAME
bitmap_remap - Apply map defined by a pair of bitmaps to another bitmap SYNOPSIS
void bitmap_remap(unsigned long * dst, const unsigned long * src, const unsigned long * old, const unsigned long * new, int bits); ARGUMENTS
dst remapped result src subset to be remapped old defines domain of map new defines range of map bits number of bits in each of these bitmaps DESCRIPTION
Let old and new define a mapping of bit positions, such that whatever position is held by the n-th set bit in old is mapped to the n-th set bit in new. In the more general case, allowing for the possibility that the weight 'w' of new is less than the weight of old, map the position of the n-th set bit in old to the position of the m-th set bit in new, where m == n % w. If either of the old and new bitmaps are empty, or if src and dst point to the same location, then this routine copies src to dst. The positions of unset bits in old are mapped to themselves (the identify map). Apply the above specified mapping to src, placing the result in dst, clearing any bits previously set in dst. For example, lets say that old has bits 4 through 7 set, and new has bits 12 through 15 set. This defines the mapping of bit position 4 to 12, 5 to 13, 6 to 14 and 7 to 15, and of all other bit positions unchanged. So if say src comes into this routine with bits 1, 5 and 7 set, then dst should leave with bits 1, 13 and 15 set. COPYRIGHT
Kernel Hackers Manual 2.6. July 2010 BITMAP_REMAP(9)
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