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PIC16877A CCS how to make the program will only printf 1?

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sfchew7

Joined: 02 Nov 2011
Posts: 8

PIC16877A CCS how to make the program will only printf 1?

Posted: Fri Nov 11, 2011 8:44 pm

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PIC16877A CCS how to make the program will only printf 1 time?
I wish to make the code below will only execute "printf" one time. How should I edit the code below ?
thanks

Code:

while(1)
{

if((input(pin_a0)==1) && (input(pin_a1)==1) && (input(pin_a2)==1))
{
delay_ms(1000);
lcd_putc("\fPL available:0");
delay_ms(1000);

printf("\fPL available:0");
delay_ms(1000);
}

else if((input(pin_a0)==0) && (input(pin_a1)==0) && (input(pin_a2)==0))
{
delay_ms(1000);
lcd_putc("\fPL available:3");
delay_ms(2000);
lcd_putc("\fPL1-empty");
lcd_putc("\nPL2-empty");
delay_ms(1000);
lcd_putc("\fPL3-empty");
delay_ms(1000);

printf("\fPL available:3");
delay_ms(1000);
}

}

Ttelmah

Joined: 11 Mar 2010
Posts: 19504

Posted: Sat Nov 12, 2011 3:09 am

Reply with quote

Something like:

Code:

//Learn to indent - makes code easier to follow....
int1 done=FALSE; //with your variable declarations

do {
if((input(pin_a0)==1) && (input(pin_a1)==1) && (input(pin_a2)==1)) {
delay_ms(1000);
lcd_putc("\fPL available:0");
delay_ms(1000);

printf("\fPL available:0");
delay_ms(1000);
done=TRUE;
}
else if((input(pin_a0)==0) && (input(pin_a1)==0) && \
(input(pin_a2)==0)) {
delay_ms(1000);
lcd_putc("\fPL available:3");
delay_ms(2000);
lcd_putc("\fPL1-empty");
lcd_putc("\nPL2-empty");
delay_ms(1000);
lcd_putc("\fPL3-empty");
delay_ms(1000);

printf("\fPL available:3");
delay_ms(1000);
done=TRUE;
}
} while (!done);

So the loop exits, when 'done' goes TRUE.

If this is the main, the chip will drop through to a hidden 'sleep' after the code, and stop running. Otherwise add whatever else you want to happen here.

Best Wishes

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