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How to store and use a number that is bigger than long in Java

4 Answers

0 votes
package javaapplication1;

import java.io.IOException;
import java.math.BigInteger;

public class JavaApplication1 {

    public static void main(String[] args) throws IOException {

        try {

            BigInteger bi = new BigInteger("999999999999999999999999999");
            System.out.println(bi);

        } catch (Exception e) {
            System.out.print(e.toString());
        }
    }
}
 
/*
           
run:
     
999999999999999999999999999
  
 */

 



answered Nov 19, 2016 by avibootz
0 votes
package javaapplication1;

import java.io.IOException;
import java.math.BigInteger;

public class JavaApplication1 {

    public static void main(String[] args) throws IOException {

        try {

            BigInteger bi = new BigInteger("999999999999999999999999999");
             System.out.println(bi.multiply(new BigInteger("3")));

        } catch (Exception e) {
            System.out.print(e.toString());
        }
    }
}
 
/*
           
run:
     
2999999999999999999999999997
  
 */

 



answered Nov 19, 2016 by avibootz
0 votes
package javaapplication1;

import java.io.IOException;
import java.math.BigInteger;

public class JavaApplication1 {

    public static void main(String[] args) throws IOException {

        try {

            BigInteger bi = new BigInteger("999999999999999999999999999");
            System.out.println(bi.add(new BigInteger("10000000")));

        } catch (Exception e) {
            System.out.print(e.toString());
        }
    }
}
 
/*
           
run:
     
1000000000000000000009999999
  
 */

 



answered Nov 19, 2016 by avibootz
edited Nov 19, 2016 by avibootz
0 votes
package javaapplication1;

import java.io.IOException;
import java.math.BigInteger;

public class JavaApplication1 {

    public static void main(String[] args) throws IOException {

        try {

            BigInteger bi = new BigInteger("999999999999999999999999999");
            System.out.println(bi.equals(new BigInteger("999999999999999999999999999")));

        } catch (Exception e) {
            System.out.print(e.toString());
        }
    }
}
 
/*
           
run:
     
true
  
 */

 



answered Nov 19, 2016 by avibootz
edited Nov 19, 2016 by avibootz
...