convert int to char to show in LCD

itoa is not implemented in CCS or in IAR.

Here's a link to various ANSI C implementations of the function

http://en.wikibooks.org/wiki/C_Programming/C_Reference/stdlib.h/itoa

Tony

i did this function to convert int to char, but it have a error, when i put float value like 24,543 he show in lcd, but when i put 33,000 for example, he don´t show anything in LCD.

void main(){
 

 InitializeLcm();
 ClearLcmScreen();
 
  float valor = 33.000;
 
  int valora, valor0, valor1, valor2, valor3, valor4;  

  if (valor < 10){
    
valora = valor * 1000;
valor0 = valora / 1000;
valor1 = valora / 100 - ( valor0 * 10);
valor2 = valora / 10 - ( (( valor0 * 10) + valor1) * 10);
valor3 = valora / 1 - ( (( valor0 * 100) + (valor1 *10) + valor2) * 10);

imprime(valor0);
PrintStr(".");
imprime(valor1);
imprime(valor2);
imprime(valor3);
PrintStr("V");

  }
 
  else if (valor > 10 ){
    
    valora = (float)valor * 1000;
    valor0 = valora / 10000;
    valor1 = valora / 1000 - ( valor0 * 10);
    valor2 = valora / 100 - ( (( valor0 * 10) + valor1) * 10);
    valor3 = valora / 10 - ( (( valor0 * 100) + (valor1 *10) + valor2) * 10);
    valor4 = valora / 1 - ( (( valor0 * 1000) + (valor1 *100) + (valor2 *10)) * 10);
    
      imprime(valor0);
      imprime(valor1);       
      PrintStr(".");      
      imprime(valor2);      
      imprime(valor3);      
      imprime(valor4);
      PrintStr("V");
  }

}

void imprime(int valor1){

if(valor1 == 0){
PrintStr("0");
}
else if(valor1 == 1){
PrintStr("1");
}
else if(valor1 == 2){
PrintStr("2");
}
else if(valor1 == 3){
PrintStr("3");
}
else if(valor1 == 4){
PrintStr("4");
}
else if(valor1 == 5){
PrintStr("5");
}
else if(valor1 == 6){
PrintStr("6");
}
else if(valor1 == 7){
PrintStr("7");
}
else if(valor1 == 8){
PrintStr("8");
}
else if(valor1 == 9){
PrintStr("9");
}

}

why this happened?

felipe paulo said:

  int valora, valor0, valor1, valor2, valor3, valor4;

Think: what is the maximum value that can be stored in an int...?

I use to employ

uint8, uint16...

to avoid having this problems :P

realmente, troquei o int para long int, deu certo.

obrigado

felipe paulo said:

valora = (float)valor * 1000;
    valor0 = valora / 10000;

Using float is slooooooow. And significantly adds to the code size too,a s it includes a lot of library code. If you don't really, really, need floats, stay away from them.
And second, divisions are slooow too. Float divisions are  (well, think of two lines of o's)

If you use the search funciton in this board, you'll find an implementation that uses subtractions increments and comparisons (which the processor natively supports) for a fast conversion.

And to answer your question about why it didn't work for 33.000: valora is an int and theiir maximum value is 32767. So your 33000 converts into a negative value, whcih you don't handle in your code.

Jens-Michael Gross said:

Using float is slooooooow. And significantly adds to the code size too,a s it includes a lot of library code. If you don't really, really, need floats, stay away from them. And second, divisions are slooow too

Indeed - if you have code space & execution time to burn, why not just use sprintf()...?!

sprintf don't work in IAR, itoa too. I change int to long int and worked

felipe paulo said:

sprintf don't work in IAR

I find that very hard to believe - sprintf is part of the standard 'C' library!

felipe paulo said:

itoa too

The reason for that has already been explained.

Yes, i know. but when i put sprintf say

Fatal Error[e89]: Too much object code produced (more than 0x1000 bytes) for this package

then, don´t work.

That's because you've filled all your code space up with the floating point stuff!

Breathe, Andy, breathe...

felipe paulo said:

but when i put sprintf say
Fatal Error[e89]: Too much object code produced (more than 0x1000 bytes) for this package