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Re: Sprintf error

I have written a routine to update an array using sprintf statement but after the execution the array doesnt seem to update. I am clueless where i have made the mistake. Can anybody help me? The program is given below.

#include<reg669.h>
#include<stdio.h>

xdata unsigned char *Arr[3];
xdata unsigned int  Val = 1;
xdata unsigned int  Val1 = 2;



void main(void)
  {
     S0CON = 0x50;   //SERIAL COM IN MODE 1
     TMOD  = 0x20;         //TIMER 1 IN AUTO RELOAD MODE
     TH1   = 0xFD;         //BAUD RATE IS 19200
     TL1   = 0x00;
     TR1   = 1;   //ENABLE TIMER 1

     Arr[0] = "LCD Testing    ";

     Arr[1] = "Programming of LCD";
     sprintf(Arr[1],"%s%u",Arr[1],Val);

     Arr[2] = "Sprintf testing ";
     sprintf(Arr[2],"%s%u",Arr[2],Val1);

     SBUF=' ';
     printf(Arr[0]);
     TI_0 = 0;
     Delay(1000);

     SBUF=' ';
     printf(Arr[1]);
     TI_0 = 0;
     Delay(1000);

     SBUF=' ';
     printf(Arr[2]);
     TI_0 = 0;
     Delay(1000);

     while(1);
  }

The 1st printf works fine whereas for the 2nd and 3rd printf statments the integer value doesnt come alont with string. I dont know the reason....

I am clueless regarding the mistake i have done.

Parents

0 Andy Neil over 17 years ago in reply to Gowri Shankar
... and think more carefully about pointers, arrays, strings, etc

xdata unsigned char *Arr[3][20];

Should be:

xdata unsigned char Arr[3][20];

ie, an array of three 20-character strings
(and don't forget that the 20 includes the NUL terminator - so that's only 19 usable characters)

Arr[0][20] = "LCD Testing ";

No, you can't do that!

You need to copy that text into the string

Arr[1][15] = "Programming of LCD"; sprintf(Arr[1][20],"%s%u",Arr[1][15],Val);

Again, you can't do that!

you need something like:

sprintf( Arr[1], "Programming of LCD %u", Val );

and you will need to repeat that each time you want to display a new value of 'Val'...

I still don't think this array of strings is helping you - you are still going to have to rebuild each string immediately before you display it - so you might just as well use printf directly!
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0 Andy Neil over 17 years ago in reply to Gowri Shankar
... and think more carefully about pointers, arrays, strings, etc

xdata unsigned char *Arr[3][20];

Should be:

xdata unsigned char Arr[3][20];

ie, an array of three 20-character strings
(and don't forget that the 20 includes the NUL terminator - so that's only 19 usable characters)

Arr[0][20] = "LCD Testing ";

No, you can't do that!

You need to copy that text into the string

Arr[1][15] = "Programming of LCD"; sprintf(Arr[1][20],"%s%u",Arr[1][15],Val);

Again, you can't do that!

you need something like:

sprintf( Arr[1], "Programming of LCD %u", Val );

and you will need to repeat that each time you want to display a new value of 'Val'...

I still don't think this array of strings is helping you - you are still going to have to rebuild each string immediately before you display it - so you might just as well use printf directly!
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0 Andy Neil over 17 years ago in reply to Andy Neil

Can't you just use something like:

void display_a_value( unsigned char value_id )
{
   switch( value_id )
   {
      case 0: printf( "this value is: %u" this_value );
              break;

      case 1: printf( "that value is: %u" that_value );
              break;

      case 2: printf( "the other value is: %u" the_other_value );
              break;

      etc...
   }
}

Of course, you'd need some meaningful identifiers, and there's room for some optimisation - but that's the basic idea.

Always start with the basics; optimise later!