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LCD output formatting

 
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pg1995



Joined: 05 Apr 2014
Posts: 31

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LCD output formatting
PostPosted: Tue Jul 22, 2014 2:08 pm     Reply with quote

Hi

I'm using scaled integers in the code which I have written to be used in a project. I was experimenting on how to output scaled integers to a LCD.

In the code I'm using millivolts and milliamperes. The current in my project won't exceed 3A (or, 3000mA) but voltage could be anywhere between 5V to 25V (i.e. 5000mV to 25000mV).

These millivolts and milliamperes can be output in volts and amperes respectively using %x.yW format where "x" is minimum field width and "y" is number of digits after the decimal point. Is this really the way to go?

Please have a look here which shows the LCD output for the code below. Let's only focus on the "if" blocks. I believe that I need to use "if" blocks so that voltages lower than 10V and over 10V could be displayed properly. Am I right?

Code:

#include <16f876.h>
#use delay(Clock=20000000)
#fuses HS, NOWDT, NOPROTECT, NOPUT, NOBROWNOUT,NOLVP

#include <Flex_LCD420.c>

void main()
{

int16 v1, v2;
signed int16 a1, a2, a3, a4;

//I need to output values in volts and amperes

   v1 = 7200;   //mV
   v2 = 14000;  //mV
   a1 = 300;    //mA
   a2 = 2000;   //mA
   a3 = 400;    //mA
 //a4 = 8000;  //mV
   a4 = 15000; //mV

lcd_init();
printf(lcd_putc, "\f");
delay_ms(500);
 
 
  while(TRUE) {
   
       
   printf(lcd_putc, "\fVi=%Lu", v1);
   printf(lcd_putc, "V");
   delay_ms(200);
   printf(lcd_putc, "  Vo=%Lu", v2);
   printf(lcd_putc, "V");
   delay_ms(200);
   printf(lcd_putc, "  \nIi=%Ld", a1);
   printf(lcd_putc, "A");
   delay_ms(200);
   printf(lcd_putc, "  Io=%Ld", a2);
   printf(lcd_putc, "A");
   printf(lcd_putc, "  \nI=%4.3w", a3);
   printf(lcd_putc, "A");
   
   if (a4 < 10000) {
      printf(lcd_putc, "  \nV=%4.3w", a4);
      printf(lcd_putc, "V");
      }
   
   if (a4 >= 10000 ) {
      printf(lcd_putc, "  \nV=%5.3w", a4);
      printf(lcd_putc, "V");
      }
       
   delay_ms(5000);
 
    }
         
}


Is there any way to display 15.000V as 15.00V? I have tried "printf(lcd_putc, " \nV=%4.3w", a4)" but it still displays 15.000V.

Thank you.
temtronic



Joined: 01 Jul 2010
Posts: 9253
Location: Greensville,Ontario

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PostPosted: Tue Jul 22, 2014 2:59 pm     Reply with quote

...... V=%4.3w......

.3w says 3 places after the decimal point

so

.2w should just give 2 places.

If you press F11 while your project is open, the CCS help files magically appear. Find the functions, then printf(). The details are in there.

hth
jay
pg1995



Joined: 05 Apr 2014
Posts: 31

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PostPosted: Tue Jul 22, 2014 3:30 pm     Reply with quote

Thanks.

I have already checked it. It gives 150.00 for V=%4.2w. So, how do I get around this? Please let me know.
Ttelmah



Joined: 11 Mar 2010
Posts: 19561

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PostPosted: Wed Jul 23, 2014 1:57 am     Reply with quote

There is no standard way to 'throw way' an extra digit at the end of %w.

1) Divide the number by 10..... Rather obvious. Smile

2) Sprintf to a string, and then throw away the last character.

3) Write your own output routine.

For #2, using strings, possibly something like:
Code:

    char temp_val[10];

    sprintf(temp_val,"%6.3Lw",a4);
    if (a4>=10000)
        temp_val[5]='\0'; //truncate the string
    printf(lcd_putc,"%s",temp_val);

Now, this is 'sneaky'....
For numbers <10000, they will fit in the 5 digits given in the output width used in the %w. Above 10000, and extra digit will be output, which I then throw away.
pg1995



Joined: 05 Apr 2014
Posts: 31

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PostPosted: Wed Jul 23, 2014 8:37 pm     Reply with quote

Thank you, Ttelmah.

Or, better still just leave it as it is and use an extra character on the LCD! Smile

Could you please also help me with the query below?

I was experimenting with the code below and was expecting the LED (the one on the left in the video) to remain on or off for at least 30 seconds but on average it remains on or off for at least 37 seconds. It would have been okay if it were 32 or 28 seconds but 37 is way too much. Where am I going wrong with it? Thanks.

Code:

#include <18f4520.h>
#device adc=10
#device *=16

#fuses H4, NOWDT, NOPUT, NOPROTECT, NOBROWNOUT,NOLVP, INTRC_IO
#use delay(Clock=32000000)

#define LED PIN_D2

void main() {

output_high(pin_d3);

while(1) {
   delay_ms(30000);
   output_toggle(LED);
   }
}


Regards
PG
Ttelmah



Joined: 11 Mar 2010
Posts: 19561

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PostPosted: Wed Jul 23, 2014 11:39 pm     Reply with quote

What clock are you actually running?.

You have both the internal RC oscillator, and an external 8MHz crystal selected.

You want one or the other.
pg1995



Joined: 05 Apr 2014
Posts: 31

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PostPosted: Thu Jul 24, 2014 2:12 am     Reply with quote

Thank you.

Ttelmah wrote:

You want one or the other.


Sorry about that. I wanted to use internal oscillator. I have updated the code but I'm getting an error now. I believe the warning can be ignored but I don't get the reason for that error. Could you please help me with it?

Warning: Line 10(1,1): Condition always TRUE
Error: Line 11(1,1): Function used but not defined: ... delay_ms SCR=836

Code:

#include <18f4520.h>
#fuses H4, NOWDT, NOPUT, NOPROTECT, NOBROWNOUT,NOLVP, INTRC_IO

#define LED pin_d2

void main() {

output_high(pin_d3);

while(1) {
   delay_ms(30000);
   output_toggle(LED);
   }
}


PS: Actually I learned how to use the internal oscillator using different past posts on this forum and one of the posts which really helped was by temtronic, the first one from top in this thread. You can see here that he is using both "INTRC_IO" fuse setting used for internal oscillator and "#use delay()" which, in my opinion, is used for external crystal. Am I missing something here?


Last edited by pg1995 on Thu Jul 24, 2014 12:04 pm; edited 2 times in total
dyeatman



Joined: 06 Sep 2003
Posts: 1938
Location: Norman, OK

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PostPosted: Thu Jul 24, 2014 11:26 am     Reply with quote

The warning can be ignored.

delay_ms() requires the #use delay line to be added
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alan



Joined: 12 Nov 2012
Posts: 357
Location: South Africa

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PostPosted: Thu Jul 24, 2014 11:52 am     Reply with quote

if you want to get rid of the warning use

while (TRUE) {

instead of

while (1) {
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