Unicode with format - macos

Question

I want to add a bunch of Emoji icons to an array. From my earlier question I found out how to write the Emoji icons in an NSString.
Now I want to make a loop and add these icons to an array. This should be fairly easy as the unicodes are in certain ranges so something like the following should do it:
for (int i = 0; i < 10; i++)
[someArray addObject:[NSString stringWithFormat:#"\U0001F43%i", i]];
Problem is, when doing so I get an error saying:
Incomplete universal character name.
Does anyone know of a way to do this?

Answer 1

That's because the escape sequence \Uxxxxxxxx is evaluated by the compiler which replaces it with the corresponding Unicode code point. Then when the method stringWithFormat: will replace the format specifier %i with the decimal representation of i. The final string is the concatenation of the characters corresponding to \Uxxxxxxxx and the characters representing i. stringWithFormat: replaces characters with other characters ; it doesn't alter existing characters.
But the problem is, here the compiler sees an incomplete escape sequence as you only wrote 7 hexadecimal digits. So it's not able to generate the string and raises an error.
The solution is to generate the character (a simple integer value) at runtime and create a string with it using +[NSString stringWithCharacters:length].
But if you look in the headers, you'll see that NSString stores its characters as unichar which is defined as an unsigned short, i.e a 16 bits-long value, whereas the Unicode code point U+1F430 (🐰) requires at least 17 bits.
So you cannot use a single unichar character to represent that code point. But don't worry: you can use two characters to represent it.
You're lost? Here the explanation! Unicode doesn't define characters, it defines code points which are arbitrary integers values in the range U+0000 – U+10FFFF. Then, the implementation decides how to represent those code point using characters. The implementation may use any data type it wants as characters as long as it manages to represent all valid code points. The simplest solution would be to use 32 bits-long integers but that would require too much memory as most of the code point you use are in the first Unicode plan (U+0000 – U+FFFF). So NSString stores the code points with the UTF-16 encoding which uses 16 bits-long characters.
In UTF-16, every code point beyond U+FFFF is stored using a pair of characters (known as a surrogate pair) in the range 0xD800 – 0xDFFF (the corresponding code points are explicitly reserved in the Unicode standard).
In conclusion, any valid Unicode code point may be represented using one or two unichar characters. The method to do so is described there. And here is a simple implementation:
static NSString *stringWithCodePoint(uint32_t codePoint)
{
// NOTE: As I edited the answer, you'll find a simpler implementation of
// this function below
unichar characters[2];
NSUInteger length;
if ( codePoint <= 0xD7FF || (codePoint >= 0xE000 && codePoint <= 0xFFFF) ) {
characters[0] = codePoint;
length = 1;
}
if ( codePoint >= 0x10000 && codePoint <= 0x10ffff ) {
codePoint -= 0x10000;
characters[0] = 0xD800 + (codePoint >> 10);
characters[1] = 0xDC00 + (codePoint & 0x3ff);
length = 2;
}
else {
length = 0; // invalid code point
}
return [NSString stringWithCharacters:characters length:length];
}
Now that we can generate a string from any valid code point, we just need to update the code to use the function we wrote before:
for (int i = 0; i < 10; i++)
[someArray addObject:stringWithCodePoint(0x0001F430 + i)];
EDIT: I just figured out a simpler method to get a NSString from a code point. It works by using -[NSString initWithBytes:length:encoding:] and the NSUTF32StringEncoding encoding:
static NSString *stringWithCodePoint(uint32_t codePoint)
{
NSString *string = [[NSString alloc] initWithBytes:&codePoint length:4 encoding:NSUTF32StringEncoding];
// You may remove the next 3 lines if you use ARC
#if ! __has_feature(objc_arc)
[string autorelease];
#endif
return string;
}

Answer 2

Note this similar question. As one of its answers explains, backslash escapes in a string literal are evaluated at compile time. If you want to make a Unicode character using a \Uxxxx escape, the xxxx all need to be numbers in the string literal.
What you can do instead, as per another answer is use the format specifier %C -- not together with the \Uxxxx escape, but on its own -- and pass in the full character code as an integer. (Actually, a wchar_t, which is a 32-bit integer on Mac OS X now, which you'll need since the character code you're looking for is more than 16 bits long.) To put this together with a base, you can just add the integers:
wchar_t base = 0x0001F430; // unfamiliar? we start with 0x for hexadecimal integers
for (int i = 0; i < 10; i++)
[someArray addObject:[NSString stringWithFormat:#"%C", base + i]];
There's also stringWithCharacters: but that explicitly takes a (16-bit) unichar, so you'd need to use a character sequence to encode your emoji in UTF-16.

Answer 3

Use %C instead of %i
so:
[someArray addObject:[NSString stringWithFormat:#"\U0001F43%C", i]];