CODEWITHSUNDEEP
javastringintegerexponential
dnl
dnl

Reputation: 365

How to convert exponential number String to an integer number String

What is the best/fastest way to convert an exponential integer number of String type (e.g. "2.4490677E7") to an integer number String (e.g. "24490677") in Java?

Edit: The input is known to always be an integer.

My current proposal is as follows (using org.apache.commons.lang.math.NumberUtils from Apache Commons):

String input = "2.4490677E7";
String res = null;
Double d = NumberUtils.toDouble(input);
if (d != 0.0d) {
   // we have a double number here
   res = String.format("%d", d.intValue());
}

Upvotes: 1

Views: 7089

Answers (4)

hi.nitish
hi.nitish

Reputation: 2974

Let me suggest the best approach. It should take care of the precision. If the string is a decimal value the precision after decimal needs to be precised.

 String str = "123456789.0111222333444555666777888999E10";
 try {
     /** DOUBLE **/
     double doubleVal = Double.parseDouble(str);//still uses 'E'
     System.out.println("Pure Double: "+ doubleVal);// Pure Double: 1.23456789011122227E18
     System.out.println(new DecimalFormat("0000000000.00000000000000").format(doubleVal));//1234567890111222270.00000000000000
     String plainStrFrmDouble = BigDecimal.valueOf(doubleVal).toPlainString();
     System.out.println("From double: " + plainStrFrmDouble);//From double: 1234567890111222270
     //convert to desired decimal format
     System.out.println(new DecimalFormat("0.0000000").format(Double.parseDouble(plainStrFrmDouble)));//1234567890111222270.0000000


     /** BIG DECIMAL **/
     String plainStrFrmBigDec = new BigDecimal(str).toPlainString();
     System.out.println("From bigdecimal: " + plainStrFrmBigDec);         //From bigdecimal:      1234567890111222333.444555666777888999
     System.out.println("From bigdecimalNonplain" + new BigDecimal(str));//From bigdecimalNonplain1234567890111222333.444555666777888999 *The most PRECISE way*
 } catch (NumberFormatException nfe) {
     System.err.println("Error while formatting.");
 }

So, I think, new BigDecimal(str) should suffice.

Upvotes: 0

Raphael Roth
Raphael Roth

Reputation: 27373

what about

Integer res = ((Double) Double.parseDouble(input)).intValue();

Upvotes: 1

Santhati Eswar
Santhati Eswar

Reputation: 61

hi we can get from this "2.4490677E7" to this "24490677" here is the code

import java.math.BigDecimal;

public class Test {
public static void main(String[] args) {

    String value = "2.4490677E7";
    BigDecimal result = new BigDecimal(value);
    System.out.println(result.longValue());
}
}

output will be : 24490677

i hope you are looking for this.. Thank you.

Upvotes: 2

Krzysztof Cichocki
Krzysztof Cichocki

Reputation: 6414

If you need precision to be untouched, then you could use BigDecimal to read the string, and then convert it to int value

Upvotes: 2

Related Questions