一个64bit的Linux Kernel占多少memory？

Question

The address space is huge for the x86-64 even though 48-bit addresses are mainly used.

On x86 32-bit machines it was pretty clear how much RAM the kernel took up. Generally around 1 GB of ZONE_NORMAL is on the bottom of memory while everything else above the 1GB in PHYSICAL (not virtual) addresses were for ZONE_HIGHMEM (for user space). This would be a 3:1 split. Of course we can have configurations were we can have a 1:3, 2:2, etc. (by changing VM_SPLIT ).

How much memory in RAM is for kernel space for 64 bit kernels?

I know the PAGE_OFFSET is set to a value far above physically addressable memory in x64 (for both 48 and 56). PAGE_OFFSET in x64 just describes the split in virtual address space, not physical (a 48 bit PAGE_OFFSET would be ffff888000000000 ).

So does 1 GB of memory house kernel space? 2GB? 3? Are there variable or macros that describe the size? Is it calculated?

Answer 1

Each user-space process can use its own 2^47 bytes (128 TiB) of virtual address space. Or more on a system with PML5 support.

The available physical RAM to back those pages is the total size of physical RAM, minus maybe 30 MiB or so that the kernel needs for its own code/data. (Not including the pagecache: Linux will use any spare pages as buffers and disk cache). This is mostly unrelated to virtual address-space limits.

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1G is how much virtual address space a kernel used up. Not how much physical RAM.

The address-space question mattered for how much memory a single process could use at the same time, but the kernel can still use all your RAM for caching file data, etc. Unless you're finding the 2^(48-1) or 2^(57-1) bytes of the low half virtual address-space range cramped, there's no equivalent problem.

See the kernel's Documentation/x86/x86-64/mm.txt for the x86-64 virtual memory map. Also Why 4-level paging can only cover 64 TiB of physical address re: x86-64 Linux not doing inconvenient HIGHMEM stuff - the entire high half of virtual address space is reserved for the kernel, and it maps all the RAM because it's a kernel.

Virtual address space usage does indirectly set a 64 TiB limit on how much physical RAM the kernel can use, but if you have less than that there's no effect. Just like how a 32-bit kernel wasn't a problem if your machine had less than 1 or 2 GiB of RAM.

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The amount of physical RAM actually reserved by the kernel depends on build options and modules, but might be something like 16 to 32 MiB.

Check dmesg output and look for something like this kernel log message from an x86-64 5.16.3-arch1 kernel I found in an old boot-log message.

Memory: 32538176K/33352340K available (14344K kernel code, 2040K rwdata, 8996K rodata, 1652K init, 4336K bss, 813904K reserved, 0K cma-reserved

Don't count the init (freed in after boot) or reserved parts; I'm pretty sure Linux doesn't actually reserve ~800 MiB in a way that makes it unusable for anything else.

Also look for the later Freeing unused decrypted memory: 2036K / Freeing unused kernel image (initmem) memory: 1652K etc. (That's the same size as the init part listed earlier, which is why you don't have to count it.)

It might also dynamically allocate some memory during startup; that initial "memory" line is just the sum of its .text , .data , and .bss sections, static code+data sizes.

Answer 2

On 64-Bit systems, the only limitation is on how much physical memory the kernel can use. The kernel will map all the available ram, and user space applications should be able to gain access to as much as the kernel can provide while maintaining sufficient for the kernel to operate.