get type of NSNumber

Question

I want to get the type of NSNumber instance.

I found out on http://www.cocoadev.com/index.pl?NSNumber this:

 NSNumber *myNum = [[NSNumber alloc] initWithB         


        

Answer 1

objCType documentation states that The returned type does not necessarily match the method the number object was created with

Secondly, other methods of comparing the class of number to a given class type or assuming boolean number instances to be shared singletons are not documented behaviour.

A more(not completely though) reliable way is to depend on NSJSONSerialisation as it correctly recognises number instances created with bool and outputs true/false in json. This is something we can expect Apple to take care of while moving with new SDKs and on different architectures. Below is the code:

+(BOOL) isBoolType:(NSNumber*) number {

    NSError* err;
    NSData* jsonData = [NSJSONSerialization dataWithJSONObject:@{@"key":number}
                                                       options:0
                                                         error:&err];

    NSString* jsonString = [[NSString alloc] 
                              initWithData:jsonData
                                  encoding:NSUTF8StringEncoding];

    return [jsonString containsString:@"true"]
            || [jsonString containsString:@"false"];

}

Answer 2

Swift Version

NSNumber is a class-cluster so each underlying type can be figured from the instance. This code avoids hard-coding the different NSNumber types by creating an instance of the expected type, and then comparing it against the unknown type.

extension NSNumber {
    var isBool: Bool {
        return type(of: self) == type(of: NSNumber(booleanLiteral: true))
    }
}

Answer 3

In Swift:

let numberType = CFNumberGetType(answer)

switch numberType {
case .charType:
    //Bool
case .sInt8Type, .sInt16Type, .sInt32Type, .sInt64Type, .shortType, .intType, .longType, .longLongType, .cfIndexType, .nsIntegerType:
    //Int
case .float32Type, .float64Type, .floatType, .doubleType, .cgFloatType:
    //Double
}

Answer 4

check object is of NSNumber type :

if([obj isKindOfClass:NSClassFromString(@"__NSCFNumber")]) { //NSNumber }

Answer 5

You can get the type this way, no string comparisons needed:

CFNumberType numberType = CFNumberGetType((CFNumberRef)someNSNumber);

numberType will then be one of:

enum CFNumberType {
   kCFNumberSInt8Type = 1,
   kCFNumberSInt16Type = 2,
   kCFNumberSInt32Type = 3,
   kCFNumberSInt64Type = 4,
   kCFNumberFloat32Type = 5,
   kCFNumberFloat64Type = 6,
   kCFNumberCharType = 7,
   kCFNumberShortType = 8,
   kCFNumberIntType = 9,
   kCFNumberLongType = 10,
   kCFNumberLongLongType = 11,
   kCFNumberFloatType = 12,
   kCFNumberDoubleType = 13,
   kCFNumberCFIndexType = 14,
   kCFNumberNSIntegerType = 15,
   kCFNumberCGFloatType = 16,
   kCFNumberMaxType = 16
};
typedef enum CFNumberType CFNumberType;

Answer 6

The reason the compiler warns you and it doesn't work is because -[NSObject className] is declared in a category on NSObject on Mac OS X (in NSScriptClassDescription.h) and not declared on iPhone. (It doesn't support AppleScript, obviously.) NSStringFromClass([myNum class]) is what you should use to be safe across all platforms. Odds are that -className is declared as a simple wrapper around NSStringFromClass() anyway...