[U-Boot-Users] still confused about configuration of memory

[U-Boot-Users] still confused about configuration of memory

[Peter Asemann]

It's embarrassing to be the fool who asks all the stupid questions, but 
I actually can't help it... The task is the following:

I have a custom board.
I've 64 MB SDRAM on CS#1 and 16MB flash on CS#0.
I want to port the U-Boot to boot Linux.

So what I read out of the README and documentation is:

1. To setup the SDRAM is my own task. My <NAME>.c of my own custom
U-Boot port implements a function initdram() that will take care of
this.
2. I have to map the SDRAM to 0x0, as the U-Boot wants me to do that.
3. As the SDRAM must be at 0x0, I must map the Flash that was there
before somewhere else.
4. In case I have "split" the U-Boot code into parts for I have other
important information in the first sectors of my flash, I can have a
CFG_MONITOR_BASE that is different from CFG_FLASH_BASE.

Are these assumptions right?

Unfortunately that was only the first question:

How does the U-Boot know which memory banks I use and which type of
memory there is? Is it possible it doesn't know?

Especially I didn't understand why one would need to preliminary map
some memory bank somewhere else. What reason could be there not to map
the memory to it's final location in the first place? The comment in the
cpu/mpc8xx/cpu_init.c (line 116 ff.) confused me even more when I was
looking for answers there.

Peter Asemann

[U-Boot-Users] still confused about configuration of memory

[Wolfgang Denk]

In message <429C7DF0.9080600@web.de> you wrote:
>
> 1. To setup the SDRAM is my own task. My <NAME>.c of my own custom
> U-Boot port implements a function initdram() that will take care of
> this.

Right.

> 2. I have to map the SDRAM to 0x0, as the U-Boot wants me to do that.

Wrong. It's not primarily U-bOot, but Linux - on  PowerPC  systems  -
which usually needs to have SDRAM starting at physical address 0.

> 3. As the SDRAM must be at 0x0, I must map the Flash that was there
> before somewhere else.

Right.

> 4. In case I have "split" the U-Boot code into parts for I have other
> important information in the first sectors of my flash, I can have a
> CFG_MONITOR_BASE that is different from CFG_FLASH_BASE.

Wrong. CFG_MONITOR_BASE is where U-Boot begins. If you  split  it  up
the base does not (necessarily) change.

> Unfortunately that was only the first question:

Wrong again. These were 4 questions already.

> How does the U-Boot know which memory banks I use and which type of
> memory there is? Is it possible it doesn't know?

You write the code to initialize the memory controller, so  you  tell
it.

> Especially I didn't understand why one would need to preliminary map
> some memory bank somewhere else. What reason could be there not to map
> the memory to it's final location in the first place? The comment in the

Assume you have more than one bank of RAM or flash, and you  want  to
map them such that they form a contiguous area of memory. To do this,
you  need  to  know  their size. To determine, you must run some test
code which probes the memory sizes. To access the memory it must have
it mapped somewhere. So what do you do? You will  have  to  use  some
preliminary mapping (at least for the second and any further banks).

Assume you have only one bank of flash  memory,  which  needs  to  be
mapped  at  the  end  of  the address space (so that the last byte of
flash memory is at 0xFFFFFFFF). In this case, you need  to  know  the
size of the device to compute the start address. ... see above.

> cpu/mpc8xx/cpu_init.c (line 116 ff.) confused me even more when I was
> looking for answers there.

Well, read the comment again, and read the memory controller  section
of the MPC8xx User's manual. Repeat until you understand.

Best regards,

Wolfgang Denk

-- 
Software Engineering:  Embedded and Realtime Systems,  Embedded Linux
Phone: (+49)-8142-66989-10 Fax: (+49)-8142-66989-80 Email: wd at denx.de
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