Verify if String is hexadecimal

Question

I have a String like \"09a\" and I need a method to confirm if the text is hexadecimal. The code I\'ve posted does something similar, it verifies that a string is a decimal

Answer 1

There's an overloaded Long.parseLong that accepts a second parameter, specifying the radix:

Long.parseLong(cadena,16);

As an alternative, you could iterate over the characters in the string and call Character.digit(c,16) on them (if any of them return -1 it's not a valid hexadecimal digit). This is especially useful if the string is too large to fit in a long (as pointed out in the comments, that would cause an exception if the first method is used). Example:

private static boolean isNumeric(String cadena) {
    if ( cadena.length() == 0 || 
         (cadena.charAt(0) != '-' && Character.digit(cadena.charAt(0), 16) == -1))
        return false;
    if ( cadena.length() == 1 && cadena.charAt(0) == '-' )
        return false;

    for ( int i = 1 ; i < cadena.length() ; i++ )
        if ( Character.digit(cadena.charAt(i), 16) == -1 )
            return false;
    return true;
}

BTW, I'd suggest separating the concerns of "testing for a valid number" and "displaying a message to the user", that's why I simply returned false in the example above instead of notifying the user first.

Finally, you could simply use a regular expression:

cadena.matches("-?[0-9a-fA-F]+");