Memory map shows more RAM than physically available

问题

I am working on a small x86 kernel. I am accessing and attempting to read the memory map that GRUB provides in the multiboot header. I have an Intel i3 cpu and 4 GiB of RAM. While running on this machine, I am reading the following memory map:

 --Base Address--          --Length--      --Type--

0x0000000000000000     0x000000000009d000     0x1
0x000000000009d000     0x0000000000003000     0x2
0x00000000000e0000     0x0000000000020000     0x2
0x0000000000100000     0x00000000bb53f000     0x1
0x00000000bb63f000     0x0000000000080000     0x2
0x00000000bb6bf000     0x0000000000100000     0x4
0x00000000bb7bf000     0x0000000000040000     0x3
0x00000000bb7ff000     0x0000000000001000     0x1
0x00000000bb800000     0x0000000004800000     0x2
0x00000000e0000000     0x0000000010000000     0x2
0x00000000feb00000     0x0000000000004000     0x2
0x00000000fec00000     0x0000000000001000     0x2
0x00000000fed10000     0x0000000000004000     0x2
0x00000000fed18000     0x0000000000002000     0x2
0x00000000fed1b000     0x0000000000005000     0x2
0x00000000fee00000     0x0000000000001000     0x2
0x00000000ffe80000     0x0000000000180000     0x2
0x0000000100000000     0x0000000038000000     0x1

When I total up the available memory areas, I get...

0x1 (Available) - 3893.8 MiB

Which seems about right, leaving the last 200ish MiB reserved for other devices. The only problem is the total of the other memory types:

0x2, 0x3, 0x4 - 331.5 MiB

Putting my total amount of RAM at 4225.3 MiB or a little over 4.1 GiB, which leads me to my questions:

Why am I totaling more than 4GiB of RAM when I only have 4GiB installed?
Why is the last base address in the memory map 0x0000000100000000? With only 4GiB of RAM, 32 bits should be the max address size needed to address all of it. Am I misunderstanding something here?

回答1:

Some thoughts:

Address space != physical memory size.
An i3 supports virtual address spaces in either 64bit, or 32bit mode with 36bit PAE (optional, with kernel support). If you actually have 4GiB of RAM available in the booted 32bit system, PAE must be enabled. How to check: https://serverfault.com/q/247080
AFAIR, those ranges could overlap and appear in any order, so that ordering and re-typing to the most restricted type or range splitting is required.
That last base address 0x0000000100000000 is >= 2^32. This is usually done, because hardware, ROM images and other special ranges are allocated below 2^32 in (all?) PCs. Thus, either PAE or long mode is required to access main memory ranges starting at or above 2^32.

Edit:

Look here for more details: http://wiki.osdev.org/Detecting_Memory_%28x86%29

Edit 2:

Today, I stumbled upon a Sysinternals tool, which shows following physical range mapping for my EFI System, without any related setting altered. As one can see, all of 64GiB main memory is mapped at 0x100000000, right at 2^32:

来源：https://stackoverflow.com/questions/18934894/memory-map-shows-more-ram-than-physically-available

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