You need to do two things: - mem remap your boot sector memory block to virtual address 0xc0008000 - using uboot commond "go xxx" to boot it from 0xc0008000, or define env variable "bootcmd", and run "boot".
On 8/25/06, Reddy Suneel-ASR125 <suneel.reddy at motorola.com> wrote: > > Hi, > We are working on MPC 8540, Linux kernel version is 2.6.10 from > Montavista. The bootloader used in Uboot and currently it loads the uImage > at physical memory address 0x0 and transfers control to it. We want to load > the kernel at a different address say 0x8000 and for this we made the > following changes. > > 1) Altered the Makefile to linked the kernel at virtual address 0xc0008000 > ( the default was 0xc000:0000) > 2) Modified Uboot to load kernel at 0x8000 instead of 0x0 > > The kernel space still starts from 0xc000:0000 > > When we transferred control to the kernel (loaded at 0x8000) we found that > the execution proceeds only till the mapping and invalidation on TLBs. We do > not know where the control goes after this as the further instructions does > not seems to get executed. Currently we do not have the provision to connect > a debugger and hence we are unable to make out what is happening. > > Can some one give us any clue as to what we might have done wrong? This is > our first experience on PowerPC. > > > Thanks®ards > Suneel > > _______________________________________________ > Linuxppc-embedded mailing list > Linuxppc-embedded at ozlabs.org > https://ozlabs.org/mailman/listinfo/linuxppc-embedded > > -------------- next part -------------- An HTML attachment was scrubbed... URL: http://ozlabs.org/pipermail/linuxppc-embedded/attachments/20060828/bd867c75/attachment.htm