A logical solution would be to use *string as mentioned by Ainar-G. This other answer details the possibilities of obtaining a pointer to a value (int64 but the same works for string too). A wrapper is another solution. Using just string An optional string means a string plus 1 specific value (or state) saying “not … Read more
If it is acceptable that newline (and other whitespace) characters are removed from both ends of the string then you can use let string = “\n\nBLA\nblub” let trimmed = string.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()) // In Swift 1.2 (Xcode 6.3): let trimmed = (string as NSString).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()) To remove leading newline/whitespace characters only you can (for example) use a regular … Read more
/// Construct a color from a hex code string, of the format #RRGGBB. Color hexToColor(String code) { return new Color(int.parse(code.substring(1, 7), radix: 16) + 0xFF000000); }
The & operator does string concatenation, that is, forces operands to be converted to strings (like calling CStr on them first). +, in its turn, forces addition if one of the expressions is numeric. For example: 1 & 2 gives you 12, whereas 1 + 2 “1” + 2 1 + “2” give you 3. … Read more
Here is a solution I found. It doesn’t seem very elegant, though: <%@ taglib prefix=”fn” uri=”http://java.sun.com/jsp/jstl/functions” %> <% pageContext.setAttribute(“newLineChar”, “\n”); %> ${fn:replace(item.comments, newLineChar, “; “)}
For replace you can use vbCrLf: Replace(string, vbCrLf, “”) You can also use chr(13)+chr(10). I seem to remember in some odd cases that chr(10) comes before chr(13).
For strings you can just write “abc” * 3, which works via StringOps and uses a StringBuffer behind the scenes. For characters I think your solution is pretty reasonable, although char.toString * n is arguably clearer. Do you have any reason to suspect the List.fill version isn’t efficient enough for your needs? You could write … Read more
In Scala 2.10 you can use string interpolation. val height = 1.9d val name = “James” println(f”$name%s is $height%2.2f meters tall”) // James is 1.90 meters tall
The first argument to parse() is the expected format. You have to change that to Date.parse(“E MMM dd H:m:s z yyyy”, testDate) for it to work. (Note you don’t need to create a new Date object, it’s a static method) If you don’t know in advance what format, you’ll have to find a special parsing … Read more