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Whether the kernel space and user space address ranges are always divided in 1:3 ratio.

For example if I have 4GB of address space and 4GB of physical RAM then the kernel gets 1GB and user space gets 3GB (assumed)!

So if I have 2GB of RAM then is it like the following?

kernel: 512MB
user space: 0-1536MB

Is this correct assumption?

phuclv
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sourav punoriyar
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    No, probably this will help http://www.makelinux.net/ldd3/chp-15-sect-1 – Alex Hoppus Feb 20 '16 at 13:45
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    Possible duplicate of [Linux 3/1 virtual address split](http://stackoverflow.com/questions/1658757/linux-3-1-virtual-address-split) – Sam Protsenko Feb 20 '16 at 13:48
  • Default split is 3G/1G user/kernel (high mem/low mem) split. You can change this via `CONFIG_VMSPLIT_*` options, but only for x86_32. For x86_64 we don't need high mem (see [this](http://unix.stackexchange.com/questions/32083/is-highmem-relevant-in-64-bit-linux-if-not-why)). Memory map for x86_64 described in [Documentation/x86/x86_64/mm.txt](https://git.kernel.org/cgit/linux/kernel/git/torvalds/linux.git/tree/Documentation/x86/x86_64/mm.txt). – Sam Protsenko Feb 20 '16 at 14:10

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No, the split is only for dividing up the virtual address space.

It just means that the address space from 0x00000000 up to 0xBFFFFFFF 'belongs' to or is available for mapping in user-space. Virtual addresses 0xC0000000 to 0xFFFFFFFF belong to the kernel.

The amount of available RAM and how it is used has nothing to do with how the virtual address space is partitioned in the Linux kernel.

FWIW, on ARM, you can configure what the split is so it doesn't HAVE to be 3:1 (user:kernel). It can be 1:3, 2:2 or 3:1. I'm assuming there is a similar option for the x86 arch.

tangrs
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  • Similar options must be available for all 32-bit architectures, because 64-bit architectures don't need that split at all. Details are [here](http://unix.stackexchange.com/questions/32083/is-highmem-relevant-in-64-bit-linux-if-not-why). – Sam Protsenko Feb 20 '16 at 14:14
  • Partitioning the virtual address space is still relevant in the x86_64 arch. I think you're referring to highmem not being necessary. – tangrs Feb 20 '16 at 14:20
  • Yes, that's what I'm saying. Partitioning scheme for x86_64 is described at `Documentation/x86/x86_64/mm.txt`, but there is no high mem there. And there is no option for x86_64 to choose high mem / low mem split either. – Sam Protsenko Feb 20 '16 at 14:24