Member Avatar for ankit.pandey3

Hi friends,

Can any one tell me the java program that how to convert Octa Decimal to HexaDecimal without using inbuild functions.
I tried a lot but not successful.
Please Please Please post the program...

Thanks in advance..

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  • Latest Post Latest Post by bguild

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Following 's multiply-and-add idea, below is a compact, clear Java implementation that converts an octal string to a hexadecimal string without using any built-in base-conversion helpers (no Integer.parseInt(..., 8), no toHexString, no new BigInteger(octal, 8)). It parses the octal digits left-to-right into an accumulator and then emits hex digits by repeated division by 16. 's point about learning is noted: this shows the algorithm so the mechanics are understood, not just a copy/paste solution.

Key points: the code validates digits (0–7), trims whitespace, supports an optional leading minus sign, and uses BigInteger so very large octal values won’t overflow. Time complexity is O(n) to parse plus O(k) to produce k hex digits.

import java.math.BigInteger;

public class OctalToHex {
    private static final BigInteger BASE8  = BigInteger.valueOf(8);
    private static final BigInteger BASE16 = BigInteger.valueOf(16);

    public static String octalToHex(String octal) {
        if (octal == null) throw new IllegalArgumentException("Null input");
        octal = octal.trim();
        if (octal.isEmpty()) throw new IllegalArgumentException("Empty input");

        boolean negative = false;
        if (octal.charAt(0) == '-') {
            negative = true;
            octal = octal.substring(1);
            if (octal.isEmpty()) throw new IllegalArgumentException("No digits after '-'");
        }

        BigInteger acc = BigInteger.ZERO;
        for (int i = 0; i < octal.length(); i++) {
            char c = octal.charAt(i);
            if (c < '0' || c > '7') throw new IllegalArgumentException("Invalid octal digit: " + c);
            acc = acc.multiply(BASE8).add(BigInteger.valueOf(c - '0'));
        }

        if (acc.equals(BigInteger.ZERO)) return "0";

        StringBuilder hex = new StringBuilder();
        while (acc.signum() > 0) {
            BigInteger[] qr = acc.divideAndRemainder(BASE16);
            int rem = qr[1].intValue();
            hex.append(rem < 10 ? (char)('0' + rem) : (char)('A' + rem - 10));
            acc = qr[0];
        }

        if (negative) hex.append('-');
        return hex.reverse().toString();
    }

    public static void main(String[] args) {
        System.out.println(octalToHex("17"));    // F
        System.out.println(octalToHex("10"));    // 8
        System.out.println(octalToHex("0"));     // 0
        System.out.println(octalToHex("-377"));  // -FF
    }
}

Troubleshooting notes:

  • Invalid characters (8, 9, letters) will throw an exception — trim input first if it may contain spaces.
  • For small numbers and slightly better speed, a long accumulator can be used instead of BigInteger, but watch for overflow.
  • Leading zeros are handled naturally; an input of "0007" returns "7".

Recommended Answers

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Member Avatar for JamesCherrill

There are lots of people here who will freely give their time to help you become the best Java programmer you can be. There's nobody here who is interested in helping you cheat or doing your homework for you.

DaniWeb Member Rules (which you agreed to when you signed up) include:
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Member Avatar for bguild

This is something you can easily do yourself! Just take your numeral one digit at a time from left to right and keep an accumulator that starts at 0. For each digit, multiply the accumulator by the base and then add the value of the digit. You can reverse the process to print out the number in whatever base you like by using % and division, but of course that will get the digits from right to left so you'd have to construct the string before writing it.

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