Manipulating and comparing floating points in java

Question

In Java the floating point arithmetic is not represented precisely. For example this java code:

float a = 1.2; 
float b= 3.0;
float c = a * b; 
if(c == 3.6){
    System.out.println("c is 3.6");
} 
else {
    System.out.println("c is not 3.6");
} 

Prints "c is not 3.6".

I'm not interested in precision beyond 3 decimals (#.###). How can I deal with this problem to multiply floats and compare them reliably?

Answer 1

I think it has nothing to do with Java, it happens on any IEEE 754 floating point number. It is because of the nature of floating point representation. Any languages that use the IEEE 754 format will encounter the same problem.

As suggested by David above, you should use the method abs of java.lang.Math class to get the absolute value (drop the positive/negative sign).

You can read this: http://en.wikipedia.org/wiki/IEEE_754_revision and also a good numerical methods text book will address the problem sufficiently.

public static void main(String[] args) {
    float a = 1.2f;
    float b = 3.0f;
    float c = a * b;
        final float PRECISION_LEVEL = 0.001f;
    if(Math.abs(c - 3.6f) < PRECISION_LEVEL) {
        System.out.println("c is 3.6");
    } else {
        System.out.println("c is not 3.6");
    }
}

Answer 2

It's a general rule that floating point number should never be compared like (a==b), but rather like (Math.abs(a-b) < delta) where delta is a small number.

A floating point value having fixed number of digits in decimal form does not necessary have fixed number of digits in binary form.

Addition for clarity:

Though strict == comparison of floating point numbers has very little practical sense, the strict < and > comparison, on the contrary, is a valid use case (example - logic triggering when certain value exceeds threshold: (val > threshold) && panic();)