Wolfgang Denk schreef:
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 ?
Hi,
Thanks for your reply.
For my processor, U0THR is defnied as: #define U0THR (*(volatile unsigned char *)(UART0_BASE_ADDR + 0x00))
UART0_BASE_ADDR is equal to 0xE000C000
The register itself represents the top of the UART0 transmit FIFO.
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.
I do not know any accessor functions. What do you mean with a "sync"?
Kind regards,
Remco Poelstra