ColdFire/68K Microcontrollers and ProcessorsのトピックRe: M52233demo board + interniche

M52233demo board + interniche

m52233demuser — Tue, 26 Jun 2007 01:14:58 GMT

Hi,

How do I send data to uart#1? I know #0 is the console port, and it works fine. But, when I send a byte to uart#1, say, by doing

uart_putc(1,'A');

to send an 'A', the system seems to lock up, that is, no console port reponse any more. What am I doing wrong? please help.

Thanks.

stuck

Re: M52233demo board + interniche

Kremer — Tue, 26 Jun 2007 05:59:35 GMT

Did you initialized the uart1?

If yes, a code snippet may point something more specific.

Re: M52233demo board + interniche

m52233demuser — Tue, 26 Jun 2007 08:42:30 GMT

Hi Kremer,

Thanks for the response. Yes. I think init happens to uart0 & uart1 upon power up. What is funny is that as long as there is no acitvity on uart0, I can do;

uart_putc(1,'A');

all I want and it won't lock up. But, it seems to lock when there are activities on both uart0 and uart1. Somethimes, it works for a while before the lock.

I would do something like;

printf("Test line......\n");

uart_putc(1,'A');

And I can see a few (maybe 5 - 10 lines) of Test line....on my HyperTerminal connected to uart0 before the program locks.

I keep checking, but, if you have more suggestions, I would appreciate it.

StillStuck

Re: M52233demo board + interniche

Kremer — Tue, 26 Jun 2007 23:34:12 GMT

Are you using the Coldfire_Lite example or did you develop your own?

Maybe printf is messing things up.

Try this instead of using printf:

static char test_string[15];

unsigned short i;

strcpy(test_string,"Test Uart0\n");

for(i=0;i<strlen(test_string);i++)

uart_putc(0,test_string[i];

uart_putc(1,'A');

If this works without lock, then printf is the villain.

Message Edited by Kremer on 2007-06-26 05:35 PM

Re: M52233demo board + interniche

m52233demuser — Fri, 29 Jun 2007 02:29:58 GMT

I am using Coldfire_Lite as is. I tried your suggestion, but, it seems that the result is the same.

What I do see while I am floundering around the ColdFire_Lite codes is that it seems uart ports are interrupt driven and both ports 0 and 1 are calling the same service routine uart_isr(). I was wondering if ports are stepping on each other while being serviced by the interrupt routine. So, what I tried is to put some forced time delay between using port0 and port1, and it seems to work much better.

I think, I need to disable one port interrupt before using the other. Does it sound reasonable?

Thank you for your interest.

Re: M52233demo board + interniche

Kremer — Fri, 29 Jun 2007 03:01:53 GMT

This Coldfire_Lite example isn´t a perfect serial example. On iuart.c, as you saw, they share the same isr for RX and TX, so when you print a string on uart 0 and try to print on the second the problem can happen. Adding a delay as you did can help on printing short strings on fast serial speed, but it will reduce the system perfomance and can´t be enough for long strings. Also, take a look on those uarts interrupt priorities and levels (on uart_init routine). You can set them up to avoid those crashings you mentioned.

But, there´s nothing like build your own application.