Re: Fw: [Sdcc-user] moving a character to sbuf

[<li...@ya...>]

Ok I did the baudrate, serial port initialisation , 
but got the same result....

Below is my code karthik.c 

# include<8051.h>
void sio_init(void)
void main(void)
{
  ACC=0x00;
  PSW=0x00;
  SP=0x64;
  sio_init();
  _asm;
       mov a,#"A"
       mov sbuf,a
       
  00003$: ljmp 00003$;

  _endasm;
}
void sio_init(void)
{
  ACC=0xF6;
  TH1=ACC;
  TMOD=0x27;
  TR1=1;
  SCON=0x50;
  PCON=0x80;
  TI=0;
  RI=0;
}

The above is the code for 9600 baudrate .

I still get '' as the output instead of 'A' . How
could i get my desired 'A' at the output ??

Am i correct ?? or should i need to make some other
modifications or additions ??

Do reply
karthik bala guru  

  

 --- Dave McGuire <mc...@ne...> wrote: > On
Monday, January 13, 2003, at 01:08 AM, mount me
> wrote:
> > THe following is my code :-
> >
> > # include<8051.h>
> > void main(void)
> > {
> >   _asm;
> >     mov a,#"A"
> >     mov sbuf,a
> >   _endasm;
> > }
> >
> > The output was '' . But i need to get A at output.
> >
> > Then i tried for a different character.
> > i changed the line mov a,#"A" in the above code
> > to mov a,#"P" . I the got the same output as '',
> but
> > i should get the P as output.
> 
>    I don't see anything in there regarding UART or
> timer (baud rate 
> generator) initialization...?
> 
>        -Dave
> 
> --
> Dave McGuire           "She's a cheek pincher.  I
> have scars."
> St. Petersburg, FL                          -Gary
> Nichols
> 
> 
> 
>
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