Hi vladimir

I think this kind of behaviour could be configurable. In case you have
described bootloader can protect its code. Actually, kernel does not
know itself how much RAM is presented. Bootloader must provide kernel
with memory configuration through ATAGs (see mem boot option).
Therefore, bootloader can reserve some RAM for itself and does not say
kernel about it. As you have already get kernel use all memory which
is provided with bootloader and it is not always an entry RAM.

In our bootloader, it seems they arent sending any ATAG_MEM as such,
but the are calling start_kernel() function, in which start_kernel is a
'type casted function pointer'
of the "kernel starting address".

if ATAG_MEM is not sent..how wil the kernel treat the available memory.

By the way ATAG_MEM has only start memory address.
How to specify that "this much memory need to be reserved" for bootloader?

On Mon, Jul 25, 2011 at 10:49 AM, Vladimir Murzin [off-list ref]wrote:

Hi sandeep!

On 7/25/11, sandeep kumar [off-list ref] wrote:

quoted

Hi all,
In Android target mobiles there is a facility called "Ramdump", where the
entire Ram image is copied to a file
as the kernel goes to panic.

Ramdump is implemented like this
1) As the panic() function is called target will be reset.
2) Bootloader boots up in specific mode, where it enables the USB driver

in

quoted

uploading mode.
3) Through USB entire RAM content is copied to a file in the host

computer.

quoted

4) This file is parsed through different tools and we can get the logs

info

quoted

wat exactly happened just before kernel panic happend.


Now here is the interesting part,
While the mobile's RAM image is being uploaded to host computer, our
bootloader is still running in the mobile's RAM.
How can it copy its own running region.

This triggered me the following questions,

1) When bootloader runs, will it run in a specific reserved region of RAM

or

quoted

it takes entire RAM?

AFAIK, it runs in specific region. For instance, for Das U-boot
bootloader you can get this with
$file ./uImage
uImage: u-boot legacy uImage, Linux-2.6.35-g6d019da-dirty, Linux/ARM,
OS Kernel Image (Not compressed), 2733288 bytes, Thu Sep  2 01:11:09
2010, Load Address: 0x80008000, Entry Point: 0x80008000, Header CRC:
0xEA95BF09, Data CRC: 0xF652AF6B

quoted

2) When it hands over the control to kernel, will it relinquish the

entire

quoted

RAM or certain part of RAM is reserved for it?
3) Does kernel use the entire RAM while running?

I think this kind of behaviour could be configurable. In case you have
described bootloader can protect its code. Actually, kernel does not
know itself how much RAM is presented. Bootloader must provide kernel
with memory configuration through ATAGs (see mem boot option).
Therefore, bootloader can reserve some RAM for itself and does not say
kernel about it. As you have already get kernel use all memory which
is provided with bootloader and it is not always an entry RAM.

quoted

4) When the kernel is running, is ther any chance that, control again be
taken back to bootloader?(coz i saw in bootloader code kernel is being
called,

after that code also some code is there,some printks and error handling
stuff)

Pls help me out here..



--
With regards,
Sandeep Kumar Anantapalli,

-- 
With regards,
Sandeep Kumar Anantapalli,
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