TheAlgorithms/Java · error · IllegalArgumentException

Incorrect binary digit: {}

Error message

Incorrect binary digit: {}

What it means

Thrown by BinaryToHexadecimal.binToHex(int) when a decimal digit of the binary input exceeds 1. Like the long-based binary converters, this method treats the int as a sequence of decimal digits that must be 0/1. Each group of 4 bits (consumed from least significant) is converted to one hex digit.

Source

Thrown at src/main/java/com/thealgorithms/conversions/BinaryToHexadecimal.java:37

    }

    /**
     * Converts a binary number to a hexadecimal number.
     *
     * @param binary The binary number to convert.
     * @return The hexadecimal representation of the binary number.
     * @throws IllegalArgumentException If the binary number contains digits other than 0 and 1.
     */
    public static String binToHex(int binary) {
        Map<Integer, String> hexMap = initializeHexMap();
        StringBuilder hex = new StringBuilder();

        while (binary != 0) {
            int decimalValue = 0;
            for (int i = 0; i < BITS_IN_HEX_DIGIT; i++) {
                int currentBit = binary % BASE_DECIMAL;
                if (currentBit > 1) {
                    throw new IllegalArgumentException("Incorrect binary digit: " + currentBit);
                }
                binary /= BASE_DECIMAL;
                decimalValue += (int) (currentBit * Math.pow(BASE_BINARY, i));
            }
            hex.insert(0, hexMap.get(decimalValue));
        }

        return !hex.isEmpty() ? hex.toString() : "0";
    }

    /**
     * Initializes the hexadecimal map with decimal to hexadecimal mappings.
     *
     * @return The initialized map containing mappings from decimal numbers to hexadecimal digits.
     */
    private static Map<Integer, String> initializeHexMap() {
        Map<Integer, String> hexMap = new HashMap<>();
        for (int i = 0; i < BASE_DECIMAL; i++) {

View on GitHub (pinned to fdfb9a395b)

Solutions

1. Validate with String.valueOf(binary).matches("[01]+") before calling.
2. Use Integer.toString(Integer.parseInt(binaryString, 2), 16) for a string-based path that preserves leading zeros.
3. Use Java binary literals (0b prefix) and Integer.toHexString() for in-code binary-to-hex conversion.

Example fix

// before
String hex = BinaryToHexadecimal.binToHex(102); // '2' digit rejected

// after
String hex = Integer.toString(Integer.parseInt("1010", 2), 16); // "a"

if (!String.valueOf(binary).matches("[01]+")) {
    throw new IllegalArgumentException("not binary digits: " + binary);
}
String hex = BinaryToHexadecimal.binToHex(binary);

static boolean isBinaryInt(int n) {
    return String.valueOf(n).matches("[01]+");
}

try {
    String hex = BinaryToHexadecimal.binToHex(n);
} catch (IllegalArgumentException e) {
    throw new DomainException("Invalid binary input: " + n, e);
}

When it happens

Trigger: Passing 102, 123, or any int containing a decimal digit 2-9. Passing a number intended as a binary string but represented as an int, losing leading zeros. Passing a hex or decimal value by mistake.

Common situations: Confusing int literal 0b1010 (Java binary literal) with the decimal 1010. Leading zeros lost when stored as int. User typos in binary input.

Related errors

• Incorrect binary digit: {}
• Input is not a valid binary number.
• Slope and intercept must be valid numbers.
• For input string: {}
• Input cannot be null

AI-assisted analysis of TheAlgorithms/Java@fdfb9a395b (2026-08-13). Data as JSON: /api/errors/edb1b6fb25218e0a. Report an issue: GitHub.