Basic question float vs double

[AlistairMonkeyFinger]

Quick general question, just came across a bug in my code becuase of this:

float myf = 80f;
float result1 = myf / 100;
float result2 = myf / 100f;

Why is result1 = 0.0 and result2 = 0.8 ? I guess it's something to do with 100 being a double and 100f being a float but i don't quite understand why ?

Cheers


Alistair

 

[sedj]

Your 100 there is an int, not a double - this is a double 100.0

I'm not sure where you are getting the 0.0 from either - this class yields an output of "0.8 0.8 0.8" - which is what you would expect ...

public class Test {


	public static void main(String args[]) throws Exception {
		float m = 80.0f;

		float m1 = (m/100);

		//float m2 = (m/100.0); // won't compile as a double is returned
		double m2 = (m/100.0);

		float m3 = (m/100.0f);
		System.err.println(m1 +" " +m2 +" " +m3);

	}

}

 

[AlistairMonkeyFinger]

Ah, ok, i think i have a casting problem...

    float f1 = 80;
    float f2 = f1 / 100;
    float f3 = 80 / 100;
    System.out.println(f1 + " " + f2 + " " + f3);

gives 80.0 0.8 0.0

80 / 100 is two ints divided and not casted automatically to floats, it should be 80f / 100f.

Cheers


Alistair

 

[jstreich]

Right, 80/100 is returned as an int, with truncation. Java's Strong Typing should actually give a compiler error -- but I guess they didn't bother because the float and int are both primatives.