How can I convert a NSString containing a number of any primitive data type (e.g. int, float, char, unsigned int, etc.)? The problem is, I don't know which number type the string will contain at runtime.

I have an idea how to do it, but I'm not sure if this works with any type, also unsigned and floating point values:

long long scannedNumber;
NSScanner *scanner = [NSScanner scannerWithString:aString];
[scanner scanLongLong:&scannedNumber]; 
NSNumber *number = [NSNumber numberWithLongLong: scannedNumber];

Thanks for the help.

Use an NSNumberFormatter:

NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterDecimalStyle;
NSNumber *myNumber = [f numberFromString:@"42"];

If the string is not a valid number, then myNumber will be nil. If it is a valid number, then you now have all of the NSNumber goodness to figure out what kind of number it actually is.

You can use -[NSString integerValue], -[NSString floatValue], etc. However, the correct (locale-sensitive, etc.) way to do this is to use -[NSNumberFormatter numberFromString:] which will give you an NSNumber converted from the appropriate locale and given the settings of the NSNumberFormatter (including whether it will allow floating point values).

For strings starting with integers, e.g., @"123", @"456 ft", @"7.89", etc., use -[NSString integerValue].

So, @([@"12.8 lbs" integerValue]) is like doing [NSNumber numberWithInteger:12].

Here's a working sample of NSNumberFormatter reading localized number NSString (xCode 3.2.4, osX 10.6), to save others the hours I've just spent messing around. Beware: while it can handle trailing blanks ("8,765.4 " works), this cannot handle leading white space and this cannot handle stray text characters. (Bad input strings: " 8" and "8q" and "8 q".)

NSString *tempStr = @"8,765.4";  
     // localization allows other thousands separators, also.
NSNumberFormatter * myNumFormatter = [[NSNumberFormatter alloc] init];
[myNumFormatter setLocale:[NSLocale currentLocale]]; // happen by default?
[myNumFormatter setFormatterBehavior:NSNumberFormatterBehavior10_4];
     // next line is very important!
[myNumFormatter setNumberStyle:NSNumberFormatterDecimalStyle]; // crucial

NSNumber *tempNum = [myNumFormatter numberFromString:tempStr];
NSLog(@"string '%@' gives NSNumber '%@' with intValue '%i'", 
    tempStr, tempNum, [tempNum intValue]);
[myNumFormatter release];  // good citizen

Thanks All! I am combined feedback and finally manage to convert from text input ( string ) to Integer. Plus it could tell me whether the input is integer :)

NSNumberFormatter * f = [[NSNumberFormatter alloc] init];
[f setNumberStyle:NSNumberFormatterDecimalStyle];
NSNumber * myNumber = [f numberFromString:thresholdInput.text];

int minThreshold = [myNumber intValue]; 

NSLog(@"Setting for minThreshold %i", minThreshold);

if ((int)minThreshold < 1 )
{
    NSLog(@"Not a number");
}
else
{
    NSLog(@"Setting for integer minThreshold %i", minThreshold);
}
[f release];

If you know that you receive integers, you could use:

NSString* val = @"12";
[NSNumber numberWithInt:[val intValue]];

What about C's standard atoi?

int num = atoi([scannedNumber cStringUsingEncoding:NSUTF8StringEncoding]);

Do you think there are any caveats?

You can also do this:

NSNumber *number = @([dictionary[@"id"] intValue]]);

Have fun!

I think NSDecimalNumber will do it:

Example:

NSNumber *theNumber = [NSDecimalNumber decimalNumberWithString:[stringVariable text]]];

NSDecimalNumber is a subclass of NSNumber, so implicit casting allowed.

NSDecimalNumber *myNumber = [NSDecimalNumber decimalNumberWithString:@"123.45"];
NSLog(@"My Number : %@",myNumber);

I wanted to convert a string to a double. This above answer didn't quite work for me. But this did: How to do string conversions in Objective-C?

All I pretty much did was:

double myDouble = [myString doubleValue];

Objective-C

(Note: this method doesn't play nice with difference locales, but is slightly faster than a NSNumberFormatter)

NSNumber *num1 = @([@"42" intValue]);
NSNumber *num2 = @([@"42.42" floatValue]);

Swift

Simple but dirty way

// Swift 1.2
if let intValue = "42".toInt() {
    let number1 = NSNumber(integer:intValue)
}

// Swift 2.0
let number2 = Int("42')

// Swift 3.0
NSDecimalNumber(string: "42.42") 

// Using NSNumber
let number3 = NSNumber(float:("42.42" as NSString).floatValue)

The extension-way This is better, really, because it'll play nicely with locales and decimals.

extension String {

    var numberValue:NSNumber? {
        let formatter = NSNumberFormatter()
        formatter.numberStyle = .DecimalStyle
        return formatter.number(from: self)
    }
}

Now you can simply do:

let someFloat = "42.42".numberValue
let someInt = "42".numberValue

You can just use [string intValue] or [string floatValue] or [string doubleValue] etc

You can also use NSNumberFormatter class:

Worked in Swift 3

NSDecimalNumber(string: "Your string") 

I know this is very late but below code is working for me.

Try this code

NSNumber *number = @([dictionary[@"keyValue"] intValue]]);

This may help you. Thanks

you can also do like this code 8.3.3 ios 10.3 support

[NSNumber numberWithInt:[@"put your string here" intValue]]

extension String {

    var numberValue:NSNumber? {
        let formatter = NumberFormatter()
        formatter.numberStyle = .decimal
        return formatter.number(from: self)
    }
}

let someFloat = "12.34".numberValue

Try this

NSNumber *yourNumber = [NSNumber numberWithLongLong:[yourString longLongValue]];

Note - I have used longLongValue as per my requirement. You can also use integerValue, longValue, or any other format depending upon your requirement.