linux启动全过程
参考: http://www.staroceans.org/e-book/linux-bootstrap-1.html
1. MBR里的内容属于grub
grub-2.02\grub-core\boot\i386\pc\boot.S 就是MBR的内容,该文件最后一句 .word GRUB_BOOT_MACHINE_SIGNATURE
GRUB_BOOT_MACHINE_SIGNATURE在boot.h 里被定义为 0xaa55
2. 现在不理解的一点是:
In the beginning of this post I wrote that the first instruction executed by the CPU is located at address 0xFFFFFFF0
, which is much larger than 0xFFFFF
(1MB). How can the CPU access this in real mode? This is in the coreboot documentation:
0xFFFE_0000 - 0xFFFF_FFFF: 128 kilobyte ROM mapped into address space
At the start of execution, the BIOS is not in RAM but in ROM.
既然实模式下内存地址只有20位,cpu只能访问到1MB的地址,那如何把超出1MB的地址进行映射呢? 据我所知,映射的地址必须在地址位能表达的范围之内。
参考:https://superuser.com/questions/988473/why-is-the-first-bios-instruction-located-at-0xfffffff0-top-of-ram
posted on 2018-07-05 23:56 longbigfish 阅读(126) 评论(0) 编辑 收藏 举报