Hi, I would like to know about the memory layout of kernel when it boots with bzImage [i.e Big kernel]. I understand the following things :- 1. Boot loader code [512 bytes] is loaded at 07c0:0000 address [i.e. 0x7c00 Physical address] using int 0x19 by the BIOS after POST thing. 2. Boot loader code [here I assume that it's Linux Kernel Legacy bootloader and not any other LILO or Grub] then moves itself to location 0x90000. 3. Then it loades the setup code at 0x90200 location. 4. After this, bool loader loades compressed kernel image [zImage] at address 0x10000 and then jumps to address 0x90200 [i.e at setup code]. 5. Setup code then moves kernel from 0x10000 to 0x1000 location overwriting firmware code previously present there. Now setup code changes real mode to protected mode and transfers control to 0x1000. 6. Now, you have compressed kernel Image at location 0x1000 whose intial code is head.S code. This code uncompresses the kernel at location 0x100000 and starts executing from there. FYI the above explaination is about zImage whose size can't be more than 512 KB. Can anyone tell me what happens in case of bzImage. What I understand is bzImage can't be loaded at location 0x10000 by the bootloader code [Explained in step 4], as we have limited space available only from 0x10000 to 0x90000 [i.e 512 KB], but generally the size of bzImage is > 512 KB. So it can't be loaded there completely. Moreover, bootloader can't load kernel at high memory [0x100000] because CPU is running in real mode. Is bzImage partially loaded at 0x10000 and then this partial bzImage is moved to 0x100000 instead of 0x1000 by setup code [after changing to real mode] ? If it is so, then there should be some code in head.S or misc.c [Which are the heads of bzImage] to load the rest portion of the bzImage, but I don't see this. Anyone please clear this doubt related to booting with bzImage. Thanks in advance. Sumit Sharma.
