How to convert an NSString into an NSNumber

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).

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 = NumberFormatter()
        formatter.numberStyle = .decimal
        return formatter.number(from: self)
    }
}

Now you can simply do:

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

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].

You can also do this:

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

Have fun!

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

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

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

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];

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];

I think NSDecimalNumber will do it:

Example:

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

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

What about C's standard atoi?

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

Do you think there are any caveats?

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

You can also use NSNumberFormatter class:

you can also do like this code 8.3.3 ios 10.3 support

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

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

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.

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

extension String {

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

let someFloat = "12.34".numberValue