Why is comparing floats inconsistent in Java?

Question

class Test{  
    public static void main(String[] args){  
        float f1=3.2f;  
        float f2=6.5f;  

        if(f1==3.2){
            System.out.println(\"         


        

Answer 1

To see what you're dealing with, you can use Float and Double's toHexString method:

class Test {
    public static void main(String[] args) {
        System.out.println("3.2F is: "+Float.toHexString(3.2F));
        System.out.println("3.2  is: "+Double.toHexString(3.2));
        System.out.println("6.5F is: "+Float.toHexString(6.5F));
        System.out.println("6.5  is: "+Double.toHexString(6.5));
    }
}
$ java Test
3.2F is: 0x1.99999ap1
3.2  is: 0x1.999999999999ap1
6.5F is: 0x1.ap2
6.5  is: 0x1.ap2

Generally, a number has an exact representation if it equals A * 2^B, where A and B are integers whose allowed values are set by the language specification (and double has more allowed values).

In this case,
6.5 = 13/2 = (1+10/16)*4 = (1+a/16)*2^2 == 0x1.ap2, while
3.2 = 16/5 = ( 1 + 9/16 + 9/16^2 + 9/16^3 + . . . ) * 2^1 == 0x1.999. . . p1.
But Java can only hold a finite number of digits, so it cuts the .999. . . off at some point. (You may remember from math that 0.999. . .=1. That's in base 10. In base 16, it would be 0.fff. . .=1.)