Is it possible to use NSArray, NSDictionary, and NSNumber “literals” in Xcode 4.3? (LLVM 4.0)

Question

Apparently, the new Objective-C literals have landed into the clang trunk, and thus lifted the shadowy veil of any NDA\'s.

My question… HOW can I,

Answer 1

I found a non-official way to do this… Using the Lumumba Framework on github, there is a whole kit'n'caboodle of Syntactic sugar categories that had the following defines… which achieve the desired effect.

#define $(...)        ((NSString *)[NSString  stringWithFormat:__VA_ARGS__,nil])
#define $array(...)   ((NSArray *)[NSArray arrayWithObjects:__VA_ARGS__,nil])
#define $set(...)     ((NSSet *)[NSSet setWithObjects:__VA_ARGS__,nil])
#define $map(...)     ((NSDictionary *)[NSDictionary dictionaryWithObjectsAndKeys:__VA_ARGS__,nil])
#define $int(A)       [NSNumber numberWithInt:(A)]
#define $ints(...)    [NSArray arrayWithInts:__VA_ARGS__,NSNotFound]
#define $float(A)     [NSNumber numberWithFloat:(A)]
#define $doubles(...) [NSArray arrayWithDoubles:__VA_ARGS__,MAXFLOAT]
#define $words(...)   [[@#__VA_ARGS__ splitByComma] trimmedStrings]
#define $concat(A,...) { A = [A arrayByAddingObjectsFromArray:((NSArray *)[NSArray arrayWithObjects:__VA_ARGS__,nil])]; }

So, basically, instead of…

NSArray *anArray = [NSArray arrayWithObjects:
    object, @"aWord", [NSNumber numberWithInt:7], nil];

It's just…

NSArray *anArray = $array(object, @"aWord", $int(7));

Ahhh, brevity.

Answer 2

Sorry, this is Xcode 4.4 only.