Cannot understand why this code work but this not

Question

I've done a simple aurduino sketch, I have this code, and this don't work:

#define voltageInput A0
int qntR;
float mr;
float qr;
float Distanza;
void setup() {
 // put your setup code here, to run once:
 Serial.begin(9600);
 pinMode(voltageInput,INPUT);
 mr=(130-20)/float(1023-205);
 qr=20-mr*205;
 Serial.print(mr,4);
}
void loop() {
 // put your main code here, to run repeatedly:
 qntR=analogRead(voltageInput);
 Distanza=mr*qntR+qr;
 Serial.print("Distanza: ");
 Serial.print(Distanza);
 Serial.print(" cm");
 Serial.println("");
}

And the I have this code:

#define voltageInput A0
int qntR;
float mr;
float qr;
float Distanza;
void setup() {
 // put your setup code here, to run once:
 Serial.begin(9600);
 pinMode(voltageInput,INPUT);
 mr=(130-20)/(1023-205);
 qr=20-mr*205;
 Serial.print(mr,4);
}
void loop() {
 // put your main code here, to run repeatedly:
 qntR=analogRead(voltageInput);
 Distanza=mr*qntR+qr;
 Serial.print("Distanza: ");
 Serial.print(Distanza);
 Serial.print(" cm");
 Serial.println("");
}

In which I have delete the float from mr=(130-20)/float(1023-205);,and if I delete the float the code stop working (mr is always equal to 0), why?

• Sim Son: What is is that doesn't work? Please include the serial output
• user12258338: it say that mr is always equal to 0

Answer 1

Your problem is that you are doing integer maths.

mr = (130-20)/(1023-205)

reduces to:

mr = 110/818

Which normally equates to:

mr = 0.135

But since all those numbers are integers the result is an integer, and is truncated at the decimal point, which means that mr is assigned:

mr = 0

You can force floating point maths by either including a decimal point in one of the numbers, or adding the suffux f:

mr = (130-20)/(1023-205.0)

or

mr = (130-20)/(1023-205f)

Or casting a portion of the sum to a float as your other code does.

• user12258338: but why, I have previusly said that mr is a float variable
• Majenko: Because you have just numbers the preprocessor does the calculation, not the compiler. The compiler just gets the number 0.
• user12258338: so if I do math with integers, the preprocessor will truncate the number in an integer?
• Majenko: If you just have literal numbers, yes.
• JRobert: The expression (130-20)/(1023-205) contains only integers so it is evaluated in integer arithmetic, and converted to float only to make the assignment to mr possible. In this case, the compiler evaluated it since the constants are all known at compile time, but it would get the same result if some of the elements were integer variables and had to be evaluated at run-time.