astral-plane

I know that Windows has supported supplemental planes since Windows XP. I have fonts which I know have characters outside the basic multilingual plane (BMP). For these characters, the Unicode codepoint consists of five hexadecimal digits. I do not know how to enter these characters in applications. Windows seems to only support keyboard entry of characters in the BMP. You can either enter a decimal number or some applications allow you to enter a four digit hexadecimal number. Can someone confirm how entry is managed? I don't care if it directly from the keyboard or application-assisted. (The

Updated question ¹ With regards to character classes, comparison, sorting, normalization and collations, what Unicode version or versions are supported by which .NET platforms? Original question I remember somewhat vaguely having read that .NET supported Unicode version 3.0 and that the internal UTF-16 encoding is not really UTF-16 but actually uses UCS-2, which is not the same. It seems, for instance, that characters above U+FFFF are not possible, i.e. consider: string s = "\u1D7D9"; // ("Mathematical double-struck digit one") and it stores the string "ᵽ9" . I'm basically looking for

I have an application that is supposed to deal with all kinds of characters and at some point display information about them. I use Qt and its inherent Unicode support in QChar, QString etc. Now I need the code point of a QChar in order to look up some data in http://unicode.org/Public/UNIDATA/UnicodeData.txt , but QChar's unicode() method only returns a ushort (unsigned short), which usually is a number from 0 to 65535 (or 0xFFFF). There are characters with code points > 0xFFFF, so how do I get these? Is there some trick I am missing or is this currently not supported by Qt/QChar? Each QChar

Regex.IsMatch( "foo", "[\U00010000-\U0010FFFF]" ) Throws: System.ArgumentException: parsing "[-]" - [x-y] range in reverse order. Looking at the hex values for \U00010000 and \U0010FFF I get: 0xd800 0xdc00 for the first character and 0xdbff 0xdfff for the second. So I guess I have really have one problem. Why are the Unicode characters formed with \U split into two chars in the string? They're surrogate pairs . Look at the values - they're over 65535. A char is only a 16 bit value. How would you expression 65536 in only 16 bits? Unfortunately it's not clear from the documentation how (or

How can I display a Unicode Character above U+FFFF using char in Java? I need something like this (if it were valid): char u = '\u+10FFFF'; Jon Skeet You can't do it with a single char (which holds a UTF-16 code unit), but you can use a String : // This represents U+10FFFF String x = "\udbff\udfff"; Alternatively: String y = new StringBuilder().appendCodePoint(0x10ffff).toString(); That is a surrogate pair (two UTF-16 code units which combine to form a single Unicode code point beyond the Basic Multilingual Plane). Of course, you need whatever's going to display your data to cope with it too..

I have a list of character range restrictions that I need to check a string against, but the char type in .NET is UTF-16 and therefore some characters become wacky (surrogate) pairs instead. Thus when enumerating all the char 's in a string , I don't get the 32-bit Unicode code points and some comparisons with high values fail. I understand Unicode well enough that I could parse the bytes myself if necessary, but I'm looking for a C#/.NET Framework BCL solution. So ... How would you convert a string to an array ( int[] ) of 32-bit Unicode code points? This answer is not correct. See @Virtlink

From Core Java , vol. 1, 9th ed., p. 69: The character ℤ requires two code units in the UTF-16 encoding. Calling String sentence = "ℤ is the set of integers"; // for clarity; not in book char ch = sentence.charAt(1) doesn't return a space but the second code unit of ℤ. But it seems that sentence.charAt(1) does return a space. For example, the if statement in the following code evaluates to true . String sentence = "ℤ is the set of integers"; if (sentence.charAt(1) == ' ') System.out.println("sentence.charAt(1) returns a space"); Why? I'm using JDK SE 1.7.0_09 on Ubuntu 12.10, if it's relevant.