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javadoublerounding
ocyalcin
ocyalcin

Reputation: 21

Java Rounding (Double)

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?

Upvotes: 0

Views: 123

Answers (2)

Emanuele Ivaldi
Emanuele Ivaldi

Reputation: 1332

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());

Upvotes: 1

Blake Yarbrough
Blake Yarbrough

Reputation: 2316

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

Upvotes: 0

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