int n = 10; char[] chars = new char[n]; Arrays.fill(chars, ‘c’); String result = new String(chars); EDIT: It’s been 9 years since this answer was submitted but it still attracts some attention now and then. In the meantime Java 8 has been introduced with functional programming features. Given a char c and the desired number … Read more
How about tail, with the -c switch. For example, to get the last 4 characters of “hello”: echo “hello” | tail -c 5 ello Note that I used 5 (4+1) because a newline character is added by echo. As suggested by Brad Koch below, use echo -n to prevent the newline character from being added.
Mac OS X’s “Keychain Access” application gives you access to the nice OS X password generator. Hit command-N and click the key icon. You get to choose password style (memorable, numeric, alphanumeric, random, FIPS-181) and choose the length. It also warns you about weak passwords.
You don’t need to escape the ‘ character in a String (wrapped in “), and you don’t have to escape a ” character in a char (wrapped in ‘).
Yes, there exists better way: def print_dict(d): new = {} for k, v in d.iteritems(): if isinstance(v, dict): v = print_dict(v) new[k.replace(‘.’, ‘-‘)] = v return new (Edit: It’s recursion, more on Wikipedia.)
Collections.addAll is what you want. Collections.addAll(myArrayList, ‘+’, ‘-‘, ‘*’, ‘^’); Another option is to pass the list into the constructor using Arrays.asList like this: List<Character> myArrayList = new ArrayList<Character>(Arrays.asList(‘+’, ‘-‘, ‘*’, ‘^’)); If, however, you are good with the arrayList being fixed-length, you can go with the creation as simple as list = Arrays.asList(…). Arrays.asList … Read more
The Chr function in VB.NET converts the integer back to the character: Dim i As Integer = Asc(“x”) ‘ Convert to ASCII integer. Dim x As Char = Chr(i) ‘ Convert ASCII integer to char.
I needed a replacement for hexadecimal (base 16) for similar reasons (e.g. for encoding a key, etc.), the best I could come up with is the following set of 16 characters, which can be used as a replacement for hexadecimal: 0 1 2 3 4 5 6 7 8 9 A B C D E … Read more
For Swift 5 see rustylepord’s answer. Update for Swift 3: let letters = CharacterSet.letters let digits = CharacterSet.decimalDigits var letterCount = 0 var digitCount = 0 for uni in phrase.unicodeScalars { if letters.contains(uni) { letterCount += 1 } else if digits.contains(uni) { digitCount += 1 } } (Previous answer for older Swift versions) A possible … Read more