Java: NumberFormatException when converting Binary to Hex

Question

I have homework to convert from BIN to HEX and I wrote the following algo code:

import java.util.Scanner;

public class BinaryToHex {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.print("Binary number: ");
        String b = input.next();
        int bin = Integer.parseInt(b);
        int arrlength = b.length();

        while (arrlength%4  !=  0){
            arrlength++;
        }

        int[] arrbin =  new int [arrlength];
        int digit = 0;
        String hex = "";
        String str;
        int conv;

        for (int i = arrlength-1; i>=0; i--){
            digit = bin%10;
            arrbin[i]=digit;
            bin = bin/10;
        }

        System.out.print("Hex value = ");

        for (int index = 0; index < arrlength; index=index+4){
            str = "" + arrbin[index] + "" + arrbin[index+1] + "" + arrbin[index+2] + "" + arrbin[index+3];
            switch(str){
            case "0000": str = "0"; break;
            case "0001": str = "1"; break;
            case "0010": str = "2"; break;
            case "0011": str = "3"; break;
            case "0100": str = "4"; break;
            case "0101": str = "5"; break;
            case "0110": str = "6"; break;
            case "0111": str = "7"; break;
            case "1000": str = "8"; break;
            case "1001": str = "9"; break;
            case "1010": str = "A"; break;
            case "1011": str = "B"; break;
            case "1100": str = "C"; break;
            case "1101": str = "D"; break;
            case "1110": str = "E"; break;
            case "1111": str = "F"; break;
            }
            System.out.print(str);
        }
    }

}

The problem is that when I am trying to convert bigger numbers it throws:

Exception in thread "main" java.lang.NumberFormatException: For input string: "10101010101010"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at BinaryToHex.main(BinaryToHex.java:9)

I know that the problem is connected with the int type, but I cannot figure out how to solve this. I tried to use long type - the result was same.

I would be grateful if you guys, can help me to correct this code, in order to work with bigger numbers.

Shawn Vader · Accepted Answer · 2014-03-25 10:30:09Z

3

In Java the int is a signed 32 bit so its range is

-2,147,483,648 to 2,147,483,647

So your value of 10101010101010 is outside of this range.

Try using something larger like a Long

long bin = Long.valueOf("10101010101010");
System.out.println(bin);

See Primitive Data Types

answered Mar 25, 2014 at 10:30

Shawn Vader

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1 Comment

Kirev Over a year ago

Thank you, it works ! That's what I've been searching for !I regret I am not allowed to vote up for you ... :|

AlexR · Accepted Answer · 2014-03-25 10:19:43Z

0

This number is indeed too big. Use Long.parseLong() and long type instead of int if you need such big numbers.

EDIT:

I have just understood that you actually want to parse binary number. So, use Integer.parseInt(str, 2). Otherwise the number is interpreted as decimal.

answered Mar 25, 2014 at 10:19

AlexR

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Comments

raj · Accepted Answer · 2014-03-25 10:31:49Z

0

I used this method for conversion

   public static String binaryToHex(String bin) {
   return String.format("%21X", Long.parseLong(bin,2)) ;
}

edited Mar 25, 2014 at 10:31

answered Mar 25, 2014 at 10:23

raj

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Comments

Harmlezz · Accepted Answer · 2014-03-25 10:39:04Z

0

I did it quick and dirty. Should work for all lengths:

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    System.out.print("Binary number: ");
    String b = input.next();
    while (b.length() % 4 != 0) b = "0" + b;
    StringBuilder builder = new StringBuilder();
    for (int count = 0; count < b.length(); count += 4) {
        String nibble = b.substring(count, count + 4);
        builder.append(Integer.toHexString(Integer.parseInt(nibble, 2)));
    }
    System.out.println(builder);
}

edited Mar 25, 2014 at 10:39

answered Mar 25, 2014 at 10:19

Harmlezz

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Comments

SubOptimal · Accepted Answer · 2014-03-25 11:53:18Z

0

If the input String is at maximum a 32 bit value there is no need to use a Long.

-- for any binary input till 32 bit lenght following will work
Integer.parseInt(yourBinaryString, 2)

-- for any binary input till 64 bit lenght following will work
Long.parseLong(yourBinaryString, 2)

-- for longer binary string input values have a look at BigInteger

answered Mar 25, 2014 at 11:53

SubOptimal

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