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

6.5 can be represented exactly in binary, whereas 3.2 can't. That's why the difference in precision doesn't matter for 6.5, so 6.5 == 6.5f.

To quickly refresh how binary numbers work:

100 -> 4

10 -> 2

1 -> 1

0.1 -> 0.5 (or 1/2)

0.01 -> 0.25 (or 1/4)

etc.

6.5 in binary: 110.1 (exact result, the rest of the digits are just zeroes)

3.2 in binary: 11.001100110011001100110011001100110011001100110011001101... (here precision matters!)

A float only has 24 bits precision (the rest is used for sign and exponent), so:

3.2f in binary: 11.0011001100110011001100 (not equal to the double precision approximation)

Basically it's the same as when you're writing 1/5 and 1/7 in decimal numbers:

1/5 = 0,2
1,7 = 0,14285714285714285714285714285714.