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Convert Octal to Decimal in Java

beginner· Numbers · Number Conversion

Problem

In an octal number each digit is worth a power of eight, so 517 in octal equals 5×64 + 1×8 + 7 = 335.

Given a string holding an octal number, print its decimal value, rejecting digits outside 0 to 7.

Input

"517"

Output

335

Java Program

Java

public class OctalToDecimal {
    public static void main(String[] args) {
        String octal = "517";

        int decimal = 0;
        for (char digit : octal.toCharArray()) {
            int value = digit - '0';
            if (value < 0 || value > 7) {
                System.out.println("Not a valid octal digit: " + digit);
                return;
            }
            decimal = decimal * 8 + value; // each step moves the earlier digits up one power of 8
        }

        System.out.println("Decimal of " + octal + ": " + decimal);
    }
}

▶ Run Code

Output

Decimal of 517: 335

Core Logic

Multiplying the running total by 8 before adding each digit places every earlier digit at the right power of eight.

How It Works

1digit - '0' converts each character to its numeric value.
2Any value outside 0 to 7 is reported as invalid, and the program stops.
3decimal * 8 + value shifts the total one octal place and adds the new digit.
4After the last digit, decimal holds the full base-10 value.

For 517: 5 → 5×8 + 1 = 41 → 41×8 + 7 = 335.

💡

Key Point: Validating each digit matters — without it, an input like "519" would silently produce a wrong number instead of an error.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: Each of the n digits is checked and folded into one running total.

Key Concepts

for-each loopinput validationHorner's method

Approach 2: Using Integer.parseInt() with Radix 8

Java

public class OctalToDecimalBuiltIn {
    public static void main(String[] args) {
        String octal = "517";
        System.out.println("Decimal of " + octal + ": " + Integer.parseInt(octal, 8));

        try {
            Integer.parseInt("519", 8); // 9 is not an octal digit
        } catch (NumberFormatException e) {
            System.out.println("519 is not valid octal");
        }
    }
}

▶ Run Code

Output

Decimal of 517: 335
519 is not valid octal

Core Logic

Passing 8 as the radix makes parseInt read the digits as octal and reject anything that isn't a valid octal digit.

How It Works

1Integer.parseInt("517", 8) returns 335.
2Integer.parseInt("519", 8) throws a NumberFormatException because 9 is not an octal digit.
3The catch block turns that exception into a readable message.

The two calls print 335 and then the invalid-input message.

💡

Key Point: Beware of Java source literals: int x = 0517; is also octal (335), a classic trap when someone pads a number with a leading zero.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: The parser reads each character once.

Key Concepts

Integer.parseInt()NumberFormatException

Interview Questions

How do you validate an octal string?

Every character must be between '0' and '7'; s.matches("[0-7]+") checks it in one call.

What is 777 octal in decimal?

511 (7 × 64 + 7 × 8 + 7).

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