How does the linux kernel manage less than 1GB physical memory?

Question

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 kern

Answer 1

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

Answer 2

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Hi, actually, I don't work on x86 hardware platform, so there may exist some technical errors in my post.

To my knowledge, the range between 0(or 16)MB - 896MB is listed specially while you have more RAM than that number, say, you have 1GB physical RAM on your board, which is called "low-memory". If you have more physical RAM than 896MB on your board, then, rest of the physical RAM is called highmem.

Speaking of your question, there are 512MiBytes physical RAM on your board, so actually, there is no 896, no highmem.

The total RAM kernel can see and also can map is 512MB.

'Cause there is 1-to-1 mapping between physical memory and kernel virtual address, so there is 512MiBytes virtual address space for kernel. I'm really not sure whether or not the prior sentence is right, but it's what in my mind.

What I mean is if there is 512MBytes, then the amount of physical RAM the kernel can manage is also 512MiBytes, further, the kernel cannot create such big address space like beyond 512MBytes.

Refer to user space, there is one different point, pages of user's application can be swapped out to harddisk, but pages of the kernel cannot.

So, for user space, with the help of page tables and other related modules, it seems there is still 4GBytes address space. Of course, this is virtual address space, not physical RAM space.

This is what I understand.

Thanks.

Answer 3

If the physical memory is less than 896 MB then the linux kernel maps upto that physical address lineraly.

For details see this.. http://learnlinuxconcepts.blogspot.in/2014/02/linux-addressing.html

Answer 4

The hardware provides a Memory Management Unit. It is a piece of circuitry which is able to intercept and alter any memory access. Whenever the processor accesses the RAM, e.g. to read the next instruction to execute, or as a data access triggered by an instruction, it does so at some address which is, roughly speaking, a 32-bit value. A 32-bit word can have a bit more than 4 billions distinct values, so there is an address space of 4 GB: that's the number of bytes which could have a unique address.

So the processor sends out the request to its memory subsystem, as "fetch the byte at address x and give it back to me". The request goes through the MMU, which decides what to do with the request. The MMU virtually splits the 4 GB space into pages; page size depends on the hardware you use, but typical sizes are 4 and 8 kB. The MMU uses tables which tell it what to do with accesses for each page: either the access is granted with a rewritten address (the page entry says: "yes, the page containing address x exists, it is in physical RAM at address y") or rejected, at which point the kernel is invoked to handle things further. The kernel may decide to kill the offending process, or to do some work and alter the MMU tables so that the access may be tried again, this time successfully.

This is the basis for virtual memory: from the point of view, the process has some RAM, but the kernel has moved it to the hard disk, in "swap space". The corresponding table is marked as "absent" in the MMU tables. When the process accesses his data, the MMU invokes the kernel, which fetches the data from the swap, puts it back at some free space in physical RAM, and alters the MMU tables to point at that space. The kernel then jumps back to the process code, right at the instruction which triggered the whole thing. The process code sees nothing of the whole business, except that the memory access took quite some time.

The MMU also handles access rights, which prevents a process from reading or writing data which belongs to other processes, or to the kernel. Each process has its own set of MMU tables, and the kernel manage those tables. Thus, each process has its own address space, as if it was alone on a machine with 4 GB of RAM -- except that the process had better not access memory that it did not allocate rightfully from the kernel, because the corresponding pages are marked as absent or forbidden.

When the kernel is invoked through a system call from some process, the kernel code must run within the address space of the process; so the kernel code must be somewhere in the address space of each process (but protected: the MMU tables prevent access to the kernel memory from unprivileged user code). Since code can contain hardcoded addresses, the kernel had better be at the same address for all processes; conventionally, in Linux, that address is 0xC0000000. The MMU tables for each process map that part of the address space to whatever physical RAM blocks the kernel was actually loaded upon boot. Note that the kernel memory is never swapped out (if the code which can read back data from swap space was itself swapped out, things would turn sour quite fast).

On a PC, things can be a bit more complicated, because there are 32-bit and 64-bit modes, and segment registers, and PAE (which acts as a kind of second-level MMU with huge pages). The basic concept remains the same: each process gets its own view of a virtual 4 GB address space, and the kernel uses the MMU to map each virtual page to an appropriate physical position in RAM, or nowhere at all.

Answer 5

osgx has an excellent answer, but I see a comment where someone still doesn't understand.

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.

Here is much of the confusion. There is virtual memory and there is physical memory. Every 32bit CPU has 4GB of virtual memory. The Linux kernel's traditional split was 3G/1G for user memory and kernel memory, but newer options allow different partitioning.

Why distinguish between the kernel and user space? - my own question

When a task swaps, the MMU must be updated. The kernel MMU space should remain the same for all processes. The kernel must handle interrupts and fault requests at any time.

How does virtual to physical mapping work? - my own question.

There are many permutations of virtual memory.

• a single private mapping to a physical RAM page.
• a duplicate virtual mapping to a single physical page.
• a mapping that throws a SIGBUS or other error.
• a mapping backed by disk/swap.

From the above list, it is easy to see why you may have more virtual address space than physical memory. In fact, the fault handler will typically inspect process memory information to see if a page is mapped (I mean allocated for the process), but not in memory. In this case the fault handler will call the I/O sub-system to read in the page. When the page has been read and the MMU tables updated to point the virtual address to a new physical address, the process that caused the fault resumes.

If you understand the above, it becomes clear why you would like to have a larger virtual mapping than physical memory. It is how memory swapping is supported.

There are other uses. For instance two processes may use the same code library. It is possible that they are at different virtual addresses in the process space due to linking. You may map the different virtual addresses to the same physical page in this case in order to save physical memory. This is quite common for new allocations; they all point to a physical 'zero page'. When you touch/write the memory the zero page is copied and a new physical page allocated (COW or copy on write).

It is also sometimes useful to have the virtual pages aliased with one as cached and another as non-cached. The two pages can be examined to see what data is cached and what is not.

Mainly virtual and physical are not the same! Easily stated, but often confusing when looking at the Linux VMM code.