Java Rounding (Double)

Question

double y1 = 0; 
double y2 = 0; 
double i = 0.025; 
double n = 2; 
double h1 = 2000; 
double h2 = 4000; 
y1 = Math.pow((1 + i), n) * h1;
y2 = Math.pow((1 + i), n) * h2;
double result = y1 + y2;
System.out.println(result);

I want the result as "6303.749999999999" but it gives me "6303.75". How can i solve it?

• Tunaki: You want 6303.749999999999?
• azurefrog: Possible duplicate of How to round a number to n decimal places in Java
• ergonaut: Isn't that the same as wanting 0.999999999999 with code like: double a = 1?
• user4668606: why would you even want to print an incorrect result?

Answer 1

correct result should be (and it is) 6303.750000000000017069679003611781820568916563219777423023367655509563911228609889292329171439632773399353027343750000

Try to give a look at the BigDecimal class.

 BigDecimal i = new BigDecimal(0.025);
 int n = 2;
 BigDecimal h1 = new BigDecimal(2000);
 BigDecimal h2 = new BigDecimal(4000);
 BigDecimal y1 = ((BigDecimal.ONE.add(i)).pow(n)).multiply(h1);
 BigDecimal y2 = ((BigDecimal.ONE.add(i)).pow(n)).multiply(h2);
 BigDecimal result = y1.add(y2);
 System.out.println(result.toEngineeringString());

Answer 2

The problem is that you are calculating the value 6303.75. if you add a statement where you subtract the value 0.000000000001 from the variable result then you will get the expected value 6303.749999999999.

The below code changes demonstrate how you can instead calculate 6303.749999999999:

public static void main(String[] args){
 double y1 = 0; 
 double y2 = 0; 
 double i = 0.025; 
 double n = 2; 
 double h1 = 2000; 
 double h2 = 4000; 
 y1 = Math.pow((1 + i), n) * h1;
 y2 = Math.pow((1 + i), n) * h2;
 double result = y1 + y2;
 result -= (double)0.000000000001; // this line is the key to calculating the expected result
 System.out.println(result);
}

Output:

6303.749999999999