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Asked: 2026-05-18T21:43:51+00:00

I’m learning the linux kernel internals and while reading Understanding Linux Kernel, quite a

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I’m learning the linux kernel internals and while reading “Understanding Linux Kernel”, quite a few memory related questions struck me. One of them is, how the Linux kernel handles the memory mapping if the physical memory of say only 512 MB is installed on my system.

As I read, kernel maps 0(or 16) MB-896MB physical RAM into 0xC0000000 linear address and can directly address it. So, in the above described case where I only have 512 MB:

  • How can the kernel map 896 MB from only 512 MB ? In the scheme described, the kernel set things up so that every process’s page tables mapped virtual addresses from 0xC0000000 to 0xFFFFFFFF (1GB) directly to physical addresses from 0x00000000 to 0x3FFFFFFF (1GB). But when I have only 512 MB physical RAM, how can I map, virtual addresses from 0xC0000000-0xFFFFFFFF to physical 0x00000000-0x3FFFFFFF ? Point is I have a physical range of only 0x00000000-0x20000000.

  • What about user mode processes in this situation?

  • Every article explains only the situation, when you’ve installed 4 GB of memory and the kernel maps the 1 GB into kernel space and user processes uses the remaining amount of RAM.

I would appreciate any help in improving my understanding.

Thanks..!

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1 Answer

  1. Editorial Team

    Not all virtual (linear) addresses must be mapped to anything. If the code accesses unmapped page, the page fault is risen.

    The physical page can be mapped to several virtual addresses simultaneously.

    In the 4 GB virtual memory there are 2 sections: 0x0… 0xbfffffff – is process virtual memory and 0xc0000000 .. 0xffffffff is a kernel virtual memory.

    • How can the kernel map 896 MB from only 512 MB ?

    It maps up to 896 MB. So, if you have only 512, there will be only 512 MB mapped.

    If your physical memory is in 0x00000000 to 0x20000000, it will be mapped for direct kernel access to virtual addresses 0xC0000000 to 0xE0000000 (linear mapping).

    • What about user mode processes in this situation?

    Phys memory for user processes will be mapped (not sequentially but rather random page-to-page mapping) to virtual addresses 0x0 …. 0xc0000000. This mapping will be the second mapping for pages from 0..896MB. The pages will be taken from free page lists.

    • Where are user mode processes in phys RAM?

    Anywhere.

    • Every article explains only the situation, when you’ve installed 4 GB of memory and the

    No. Every article explains how 4 Gb of virtual address space is mapped. The size of virtual memory is always 4 GB (for 32-bit machine without memory extensions like PAE/PSE/etc for x86)

    As stated in 8.1.3. Memory Zones of the book Linux Kernel Development by Robert Love (I use third edition), there are several zones of physical memory:

    • ZONE_DMA – Contains page frames of memory below 16 MB
    • ZONE_NORMAL – Contains page frames of memory at and above 16 MB and below 896 MB
    • ZONE_HIGHMEM – Contains page frames of memory at and above 896 MB

    So, if you have 512 MB, your ZONE_HIGHMEM will be empty, and ZONE_NORMAL will have 496 MB of physical memory mapped.

    Also, take a look to 2.5.5.2. Final kernel Page Table when RAM size is less than 896 MB section of the book. It is about case, when you have less memory than 896 MB.

    Also, for ARM there is some description of virtual memory layout: http://www.mjmwired.net/kernel/Documentation/arm/memory.txt

    The line 63 PAGE_OFFSET high_memory-1 is the direct mapped part of memory

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