Double value with specific precision in java

Question

I'm programming a simple java program. I need to get a string from input and divide it into two parts: 1-double 2-string. Then I need to do a simple calculation on the double and send the result to the output with specific precision(4). It works fine, but there is a problem when the input is 0, then it doesn't work properly.

For example for these input, output will be:

1 kg
output:2.2046

3.1 kg
output:6.8343

But when the input is 0, the output should be 0.0000, but it shows 0.0 . What should I do to force it to show 0.0000?

I read similar post about double precision, they suggest something like BigDecimal class, but I can't use them in this case, my code for doing this is:

line=input.nextLine();
array=line.split(" ");
value=Double.parseDouble(array[0]);
type=array[1];
value =value*2.2046;
String s = String.format("%.4f", value);
value = Double.parseDouble(s);
System.out.print(value+" kg\n");

Solution 1

DecimalFormat will allow you to define how many digits you want to display. A '0' will force an output of digits even if the value is zero, whereas a '#' will omit zeros.

System.out.print(new DecimalFormat("#0.0000").format(value)+" kg\n"); should to the trick.

See the documentation

Note: if used frequently, for performance reasons you should instantiate the formatter only once and store the reference: final DecimalFormat df = new DecimalFormat("#0.0000");. Then use df.format(value).

Solution 2

add this instance of DecimalFormat to the top of your method:

DecimalFormat four = new DecimalFormat("#0.0000"); // will round and display the number to four decimal places. No more, no less.

// the four zeros after the decimal point above specify how many decimal places to be accurate to.
// the zero to the left of the decimal place above makes it so that numbers that start with "0." will display "0.____" vs just ".____" If you don't want the "0.", replace that 0 to the left of the decimal point with "#"

then, call the instance "four" and pass your double value when displaying:

double value = 0;
System.out.print(four.format(value) + " kg/n"); // displays 0.0000

Solution 3

System.out.format("%.4f kg\n", 0.0d) prints '0.0000 kg'

Solution 4

I suggest you to use the BigDecimal class for calculating with floating point values. You will be able to control the precision of the floating point arithmetic. But back to the topic :)

You could use the following:

static void test(String stringVal) {
    final BigDecimal value = new BigDecimal(stringVal).multiply(new BigDecimal("2.2046"));
    DecimalFormat df = new DecimalFormat();
    df.setMaximumFractionDigits(4);
    df.setMinimumFractionDigits(4);
    System.out.println(df.format(value) + " kg\n");
}

public static void main(String[] args) {
    test("0");
    test("1");
    test("3.1");
}

will give you the following output:

0,0000 kg

2,2046 kg

6,8343 kg

View more solutions

Comments

[Raedwald]

See also stackoverflow.com/questions/10959424/… but this is not a duplicate of that because of the formatting wanted for 0.

[Aaron]

BlackLight's answer looks better.

[Blacklight]

I already did, replace your last line which prints out your value.

[Mehrdad Salimi]

i used your suggestion, but it doesn't work, as you said BlackLight's answer is what i want, Thanks,

[Mehrdad Salimi]

it worked for me, but please change it to ("#0.0000"), your answer's out put is .0000 , but if we use ("#0.0000") , it show 0.0000, thanks for your help.

[Aaron]

It's impossible for my answer to "not work" :D It's just adding "0"'s to a string. It is a sub-optimal solution though. No point re-inventing wheels!

[Mehrdad Salimi]

i think it doesn't work because of this part value=Double.parseDouble(array[0]);

[Aaron]

@MehrdadSalimi I meant add the padding on the resultant String :) That's what (i presume) decimalformat will be doing internally.

[Mehrdad Salimi]

LOL, Sorry i have english problem, :)

[Mehrdad Salimi]

Thanks for your info, i didn't know how use bigDecimal to set precision.