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    <p>Not all virtual (linear) addresses must be mapped to anything. If the code accesses unmapped page, the page fault is risen. </p> <p>The physical page can be mapped to several virtual addresses simultaneously.</p> <p>In the 4 GB virtual memory there are 2 sections: 0x0... 0xbfffffff - is process virtual memory and 0xc0000000 .. 0xffffffff is a kernel virtual memory.</p> <ul> <li>How can the kernel map 896 MB from only 512 MB ?</li> </ul> <p>It maps up to 896 MB. So, if you have only 512, there will be only 512 MB mapped.</p> <p>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).</p> <ul> <li>What about user mode processes in this situation?</li> </ul> <p>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.</p> <ul> <li>Where are user mode processes in phys RAM?</li> </ul> <p>Anywhere.</p> <ul> <li>Every article explains only the situation, when you've <em>installed 4 GB of memory</em> and the </li> </ul> <p>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)</p> <p>As stated in <code>8.1.3. Memory Zones</code> of the book <code>Linux Kernel Development</code> by Robert Love (I use third edition), there are several zones of physical memory:</p> <ul> <li>ZONE_DMA - Contains page frames of memory below 16 MB</li> <li>ZONE_NORMAL - Contains page frames of memory at and above 16 MB and below 896 MB</li> <li>ZONE_HIGHMEM - Contains page frames of memory at and above 896 MB</li> </ul> <p>So, if you have 512 MB, your ZONE_HIGHMEM will be empty, and ZONE_NORMAL will have 496 MB of physical memory mapped.</p> <p>Also, take a look to <code>2.5.5.2. Final kernel Page Table when RAM size is less than 896 MB</code> section of the book. It is about case, when you have less memory than 896 MB.</p> <p>Also, for ARM there is some description of virtual memory layout: <a href="http://www.mjmwired.net/kernel/Documentation/arm/memory.txt" rel="noreferrer">http://www.mjmwired.net/kernel/Documentation/arm/memory.txt</a></p> <p>The line 63 <code>PAGE_OFFSET high_memory-1</code> is the direct mapped part of memory</p>
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    1. POHow does the linux kernel manage less than 1GB physical memory?
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    1. POHow does the linux kernel manage less than 1GB physical memory?
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    1. COosgx, based on what you mentioned, "So, if you have 512 MB, your ZONE_NORMAL will have 496 MB of physical memory mapped." if that is the case, there will not remain any place for a user-space page in the physical memory. Unless, ZONE_NORMAL pages can be swapped out? Look forward to what you think.
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    2. CO@TheLoneJoker, Kernel can allocate phys pages for user space process both from phys. zones ZONE_NORMAL and ZONE_HIGHMEM. (e.g. as at http://linux-mm.org/HighMemory for 2G system: "It is important that the kernel allocates process and page cache memory from both zones") If ZONE_HIGHMEM is empty, kernel will allocate page from ZONE_NORMAL.
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    3. CO@TheLoneJoker, The swapping is a process on VIRTUAL Page, or, you can name it "unmapping virtual page from physical". So, any user page from virtual address space can be unmapped and swapped out.
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