user7328: I am reading a string which i have to convert into a float with a maximum of precision.I didn't want to use toFloat() function since it make a truncation so i made my own code.
My problem is that the final result is also approximated.
I need a precision of 4 digits after the float.I read the documentation about float in the arduino website.They say it should be maximum 7 digits of precision but i don't get it.When i test my code on the value of '23,459' ,all i get is 23,46.
Any suggestions?
void setup() { 

 Serial.begin(9600); 


} 
int longu=0;
char data='2';
  String data2="23,459";
  float val=0;
  int posVir=0;
  int t=0;
  float postVir=0;
  float preVir=0;
  int lonPostVir=0;
  int lonPreVir=0;
  float posvir2=0;
  void loop() 
{


   lonPostVir=(data2.substring(posVir+1)).length();
   lonPreVir=(data2.substring(0,posVir)).length();

   longu=data2.length();
   posVir=data2.indexOf(",");
   preVir=(data2.substring(0,posVir)).toFloat();
   postVir=(data2.substring(posVir+1)).toFloat();
   posvir2=postVir*pow(10,-lonPostVir);
   val=preVir+postVir*pow(10,-lonPostVir);

  Serial.println(val);

    } 


user20785: Just write the amount of numbers you want to get float here Serial.println(val); after the val.
example:
Serial.println(val, 5);
It will give 5 digits after the decimal point

user1585: The simplest way to get more decimal places to print is this:
Serial.println(val,NUMBER_OF_DECMIAL_PLACES);

This will print out any number of digits after the decimal place. Regular print without the number of digits specified will default to 2 decimal places. for example, here is some code and the result that gets printed:
float val = 23.459;
Serial.println(val);
Serial.println(val, 4);
Serial.println(val, 10);


23.46
23.4590
23.4589996337

You may notice that with 10 decimal places printed you can start to see how much the actual value of 23.459 needs to be approximated to fit within a floating point number. This is similar to how 1/3 can not be represented 100% accurately in a finite number of decimal digits; 459/1000 cannot be represented 100% accurately in a finite number of binary bits, so it is approximated.
Also, note that for non-floating point numbers the optional second argument to print and println is the base it should be printed in (dec, hex, oct, ...) and has nothing to do with precision.

7328: <p>I am reading a string which i have to convert into a float with a maximum of precision.I didn't want to use toFloat() function since it make a truncation so i made my own code.
My problem is that the final result is also approximated.
I need a precision of 4 digits after the float.I read the documentation about float in the arduino website.They say it should be maximum 7 digits of precision but i don't get it.When i test my code on the value of '23,459' ,all i get is 23,46.
Any suggestions?</p>

<pre><code>void setup() { 

 Serial.begin(9600); 


} 
int longu=0;
char data='2';
  String data2="23,459";
  float val=0;
  int posVir=0;
  int t=0;
  float postVir=0;
  float preVir=0;
  int lonPostVir=0;
  int lonPreVir=0;
  float posvir2=0;
  void loop() 
{


   lonPostVir=(data2.substring(posVir+1)).length();
   lonPreVir=(data2.substring(0,posVir)).length();

   longu=data2.length();
   posVir=data2.indexOf(",");
   preVir=(data2.substring(0,posVir)).toFloat();
   postVir=(data2.substring(posVir+1)).toFloat();
   posvir2=postVir*pow(10,-lonPostVir);
   val=preVir+postVir*pow(10,-lonPostVir);

  Serial.println(val);

    } 
</code></pre>
 20785: <p>Just write the amount of numbers you want to get float here Serial.println(val); after the val.
example:
Serial.println(val, 5);
It will give 5 digits after the decimal point</p>
 1585: <p>The simplest way to get more decimal places to print is this:</p>
<pre><code>Serial.println(val,NUMBER_OF_DECMIAL_PLACES);
</code></pre>
<p>This will print out any number of digits after the decimal place. Regular print without the number of digits specified will default to 2 decimal places. for example, here is some code and the result that gets printed:</p>
<pre><code>float val = 23.459;
Serial.println(val);
Serial.println(val, 4);
Serial.println(val, 10);
</code></pre>
<blockquote>
<p>23.46</p>
<p>23.4590</p>
<p>23.4589996337</p>
</blockquote>
<p>You may notice that with 10 decimal places printed you can start to see how much the actual value of 23.459 needs to be approximated to fit within a floating point number. This is similar to how 1/3 can not be represented 100% accurately in a finite number of decimal digits; 459/1000 cannot be represented 100% accurately in a finite number of binary bits, so it is approximated.</p>
<p>Also, note that for non-floating point numbers the optional second argument to <code>print</code> and <code>println</code> is the base it should be printed in (dec, hex, oct, ...) and has nothing to do with precision.</p>
