Converting a large binary string to hexadecimal

Question

I have a binary string:

1010010111100101100010101010011011010001111100000010101000000000010000000111110111100"

How can I convert it to a hex string?

I tried with the wrapper classes Long and Integer, but it didn’t work for me, throwing a NumberFormatException.

Calculate the decimal value then use Integer.toHexString(value); — HectorLector
– HectorLector, Commented Jun 4, 2013 at 12:56
possible duplicate of Translating a String containing a binary value to Hex - This was the very first Google result using java Convert Binary String to Hex String — John B
– John B, Commented Jun 4, 2013 at 12:56
@JohnB The answer there refers to using int / long, which are too small in this instance. — Michael Berry
– Michael Berry, Commented Jun 4, 2013 at 13:01

Michael Berry · Accepted Answer · 2013-06-04 13:08:41Z

11

You need to use BigInteger for this - the number is too big to fit in a primitive value. The biggest number that can be stored in a long is 9223372036854775807, whereas the equivalent value in decimal of this binary string is much bigger, 25069592479040759763832764. That's why you're getting the NumberFormatException.

So with BigInteger:

String s = "1010010111100101100010101010011011010001111100000010101000000000010000000111110111100";
BigInteger b = new BigInteger(s, 2);
System.out.println(b.toString(16));

...which gives:

14bcb154da3e0540080fbc

edited Jun 4, 2013 at 13:08

answered Jun 4, 2013 at 12:56

Michael Berry

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1 Comment

David L Over a year ago

WARNING - if you use the same method for binary strings that may begin with zeros, they may be dropped when converting to hex.

bowmore · Accepted Answer · 2013-06-04 13:05:44Z

2

As the length of your binary String may exceed the capacity of an Integer or Long, it's best to use BigInteger. Remember that in Java an int is always 32 bits, and a long 64 bits.

String binaryString = "1010010111100101100010101010011011010001111100000010101000000000010000000111110111100";
String hexString = new BigInteger(binaryString, 2).toString(16);

edited Jun 4, 2013 at 13:05

answered Jun 4, 2013 at 13:00

bowmore

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Comments

Muhammad Sajjad · Accepted Answer · 2016-08-08 11:32:59Z

2

public static String convertBinaryToHex(String binInPut) {
    int chunkLength = binInPut.length() / 4, startIndex = 0, endIndex = 4;
    String chunkVal = null;
    for (int i = 0; i < chunkLength; i++) {
        chunkVal = binInPut.substring(startIndex, endIndex);
         System.out.println(Integer.toHexString(Integer.parseInt(chunkVal, 2)));
        startIndex = endIndex;
        endIndex = endIndex + 4;
    }

    return binInPut;

}

answered Aug 8, 2016 at 11:32

Muhammad Sajjad

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Paul Vargas · Accepted Answer · 2013-06-04 13:00:23Z

1

If you are using big numbers:

String hexString = new BigInteger(binaryString, 2).toString(16);

answered Jun 4, 2013 at 13:00

Paul Vargas

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1 Comment

Zartc Over a year ago

BigInteger will trim leading zeros. This can be a problem if the bin string start with "00"

Govind yadav · Accepted Answer · 2023-01-25 07:54:26Z

0

I use this

fun String.binToHex(): String {
    val out = StringBuilder()
    val outArray = this.deviceInParts(4)
    outArray.forEach {
        if (it == "0000") out.append('0')
        if (it == "0001") out.append('1')
        if (it == "0010") out.append('2')
        if (it == "0011") out.append('3')
        if (it == "0100") out.append('4')
        if (it == "0101") out.append('5')
        if (it == "0110") out.append('6')
        if (it == "0111") out.append('7')
        if (it == "1000") out.append('8')
        if (it == "1001") out.append('9')
        if (it == "1010") out.append('A')
        if (it == "1011") out.append('B')
        if (it == "1100") out.append('C')
        if (it == "1101") out.append('D')
        if (it == "1110") out.append('E')
        if (it == "1111") out.append('F')
    }
    return out.toString()
}

fun String.deviceInParts(parts: Int): ArrayList<String> {
    val outArray = arrayListOf<String>()
    for (i in 0 until this.length step parts) {
        outArray.add(this.subString(i, parts))
    }
    return outArray
}

answered Jan 25, 2023 at 7:54

Govind yadav

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