Linux uses a virtual memory system where all of the addresses are virtual addresses and not physical addresses. These virtual addresses are converted into physical addresses by the processor. To make this translation easier, virtual and physical memory are divided into pages. Each of these pages is given a unique number; the page frame number. Some page sizes can be 2 KB, 4 KB, etc. But how is this page size number determined? Is it influenced by the size of the architecture? For example, a 32-bit bus will have 4 GB address space.
how is page size determined in virtual address space?
linux;kernel;memory;hardware
You can find out a system's default page size by querying its configuration via the getconf command:$ getconf PAGE_SIZE4096or $ getconf PAGESIZE4096NOTE: The above units are typically in bytes, so the 4096 equates to 4096 bytes or 4kB.This is hardwired in the Linux kernel's source here:Example$ more /usr/src/kernels/3.13.9-100.fc19.x86_64/include/asm-generic/page.h....../* PAGE_SHIFT determines the page size */#define PAGE_SHIFT  12#ifdef __ASSEMBLY__#define PAGE_SIZE   (1 << PAGE_SHIFT)#else#define PAGE_SIZE   (1UL << PAGE_SHIFT)#endif#define PAGE_MASK   (~(PAGE_SIZE-1))How does shifting give you 4096?When you shift bits, you're performing a binary multiplication by 2. So in effect a shifting of bits to the left (1 << PAGE_SHIFT) is doing the multiplication of 2^12 = 4096.$ echo 2^12 | bc4096