[U-Boot] Strange data behaviour

[Wolfgang Denk]

Dear Remco Poelstra,

In message <499AA5B8.7030805 at duran-audio.com> you wrote:
> 
> I'm trying to get my LPC2468 based board to work. I've some problems 
> with the external memory databus.
> I would like to know the settings of internal registers to see whether 
> I've initialized them correctly, so I tried making a function that 
> prints the content over the serial link. I'm still in the lowlevel_init 
> function, so I do not have the U-boot puts functions available.
> I wrote the following function to convert a long to a HEX string:
> -------------------
> void print_long(unsigned long data) {
>    char i;
>    char 
> hex_data[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
> 
>    for(i=0; i<8; i++) {
>      while((U0LSR & (1<<5)) == 0); /* Wait for empty U0THR */
> //    U0THR = hex_data[(0xDEADBEEF>>i*4)&0xF];

For clearity of the code I would write that as "...>>(i*4)...", but
that's not relevant here.

>      U0THR = hex_data[(0xDEADBEEF>>4)&0xF];
>    }
> }
> ---------------
> When I run it likes this I get 8 E's. Which is what I expect. When I run 
> it with the commented-out line, I get back 8 0x0's. So it seems that the 
> output is only correct when it is constant.
> Does anyone have a clue on why that is?

What exactly is U0THR ? 

The only definition I can find in the U-Boot sources is here:

include/asm-arm/arch-lpc2292/lpc2292_registers.h:#define U0THR 0xE000C000

but that is obviously not what you are using.

You probably forget the effects of compiler optimization and/or
caching here.


I bet you are using a plain  pointer  access  without  a  "volatile",
which is essential here. Maybe even some form of "sync" is needed. In
any  case,  you  should  use appropriate accessor functions to access
device registers.

Best regards,

Wolfgang Denk

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