CODEWITHSUNDEEP
javadecimalrounding
max
max

Reputation: 65

Round according to nearest decimal value - java

I have used this one

Math.round(input * 100.0) / 100.0;

but it didn't work with my requirement. I want to round into two decimal points according to nearest decimal value.

firstly I want to check fourth decimal value if it is 5 or above I want to add 1 into 3rd decimal value then want to check 3rd one if it is 5 or above I want to add 1 into 2nd one

ex: if I have 22.3246

refer above example number. 4th decimal no is 6. we can add 1 into 3rd decimal value.

result : 22.325

now 3rd decimal no is 5. we can add 1 into 2nd decimal value

result : 22.33

I want to get result 22.33

Upvotes: 2

Views: 127

Answers (3)

Ralf Kleberhoff
Ralf Kleberhoff

Reputation: 7290

According to your requirement, the greatest number to be rounded down to 22.32 is 22.3244444444.... From 22.32444...4445 on, iteratively rounding digit per digit will lead to 22.33.

So, you can add a bias of 0.005 - 0.0044444444 (standard rounding threshold minus your threshold), being 5/9000 and then round to the next 1/100 the standard way:

    double BIAS = 5.0 / 9000.0;
    double result = Math.round((input + BIAS) * 100.0) / 100.0;

Upvotes: 0

dems98
dems98

Reputation: 884

You can use BigDecimal class to accomplish this task, in combination with a scale and RoundingMode.HALF_UP.

Sample code:

System.out.println(new BigDecimal("22.3246").divide(BigDecimal.ONE,3,RoundingMode.HALF_UP)
                    .divide(BigDecimal.ONE,2,RoundingMode.HALF_UP));

Output:

22.33

Upvotes: 0

dyon048
dyon048

Reputation: 174

If you want to respect the 4th decimal digit, you can do it in this way:

double val = Math.round(22.3246 * 1000.0) / 1000.0;
double result = Math.round(val * 100.0) / 100.0;
System.out.println(result); // print: 22.33

Upvotes: 3

Related Questions