How to convert an NSString into an NSNumber

Question

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.

Dave DeLong · Accepted Answer · 2020-08-24T20:52:40.567

1267

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.

edited Aug 24 '20 at 20:52

answered Sep 19 '09 at 16:21

Dave DeLong

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For people where it doesn't seem to work: Check if it's related to your locale. The NSNumberFormatter (as far as I know) by default uses the US locale, i.e. expects the decimal separator to be the "." character. If you use "," to separate the fraction, you may need to tell the formatter to use your current locale: [f setLocale:[NSLocale currentLocale]]; – pille Sep 28 '12 at 09:50
2

@pille by default the locale is the `+currentLocale`, but you're correct; one must always consider the locale when dealing with converting stuff to-and-from human-readable form. – Dave DeLong Nov 22 '12 at 05:57
Where did you find that it was currentLocale by default? I can confirm that numberFromString is using dot notation even on a French phone where comma is used. I can confirm that by default, when you output a number, it uses the current locale by default – allaire Jul 24 '15 at 03:19
But what if the string can contain either int or float? – GeneCode Dec 09 '16 at 07:40

score 183 · Answer 2 · edited Sep 19 '09 at 16:24

183

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).

edited Sep 19 '09 at 16:24

Dave DeLong

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answered Sep 19 '09 at 16:18

Barry Wark

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+q Depending on the situation, non-locale-sensitive might actually be the correct way. – Thilo Sep 12 '11 at 06:46
2

I had to convert `@"2000"` to an `int` and `[@"2000" integerValue]` worked nicely and is a little simpler for my case. – Besi Feb 02 '12 at 15:09
2

There are also huge performance differences among these methods. – Tom Andersen Jun 21 '13 at 14:56
2

this does not work with ARC: it won't convert NSInteger to NSNumber. I have to further use `[NSNumber numberWithInteger ...]` – learner Aug 03 '14 at 03:08
What @Thilo said. "Correct" and "locale-sensitive" are not synonyms. The correct thing to do actually depends upon how the number got into the string in the first place. If you read it from user input, then local-sensitive is what you want. From any other source (i.e. you put it there programmatically, you're consuming it as part of the response from some external API, etc.) local-sensitive is not appropriate and could even give incorrect results. – aroth Mar 09 '16 at 04:43

Kevin R · Answer 3 · 2023-06-01T13:02:31.520

147

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

edited Jun 01 '23 at 13:02

answered Dec 09 '13 at 13:25

Kevin R

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I prefer this based on profiling when I was parsing a large amount of string values. Using this syntax rather then an NSNumberFormatter led to significant reduction in time spent parsing the string to NSNumber. And yes the NSNumberFormatter was cached and reused. – androider Mar 05 '14 at 12:27
3

this literal syntax didn't exist when this question was asked. I'd say this is the correct answer nowadays though. – Chris Sep 25 '14 at 22:09
1

I agree, but do note that some other countries use the `,` and `.` the other way around (eg `42.000,42`. Something `floatValue` probably does not account for? – Kevin R Oct 27 '14 at 12:49
1

This is the nicest solution on here, however I have just run a test and you will lose unsigned precision so WATCH OUT!!! – Stoff81 Apr 28 '15 at 11:47
In my use case, I needed to get an NSNumber dictionary key from an NSString, so `@([@"42" intValue])` worked perfectly. Not worried about Locale. – Chuck Krutsinger Aug 18 '17 at 17:13

ma11hew28 · Answer 4 · 2014-04-25T16:59:51.603

87

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].

edited Apr 25 '14 at 16:59

answered Sep 29 '10 at 13:04

ma11hew28

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score 51 · Answer 5 · answered Jun 05 '13 at 13:13

51

You can also do this:

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

Have fun!

answered Jun 05 '13 at 13:13

MacMark

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This method not converting all number of like 123,45. Just converting 123. – yasin89 Jan 23 '23 at 07:41

score 26 · Answer 6 · answered Apr 10 '12 at 00:35

26

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

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

answered Apr 10 '12 at 00:35

Ad1905

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miker · Answer 7 · 2010-11-01T20:05:53.040

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

this doesn't work for me. I'm developing for 5.0 with xcode 4.3.2. any ideas why? — acecapades, Sep 07 '12 at 03:34

score 5 · Answer 8 · edited May 23 '17 at 12:18

5

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];

edited May 23 '17 at 12:18

Community

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answered Sep 14 '13 at 14:39

Anthony

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This doesn't convert the NSInteger to a NSNumber. It converts it to a double. – FabKremer Mar 10 '15 at 18:10

score 5 · Answer 9 · edited Jul 30 '16 at 15:43

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];

score 4 · Answer 10 · answered Jul 20 '13 at 08:57

4

I think NSDecimalNumber will do it:

Example:

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

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

answered Jul 20 '13 at 08:57

JeanNicolas

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score 2 · Answer 11 · answered Jul 27 '12 at 12:55

2

What about C's standard atoi?

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

Do you think there are any caveats?

answered Jul 27 '12 at 12:55

martin

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have you read the question? you're to provide an NSNumber instance (an Objective-C object) not an int primitive. – Motti Shneor Dec 31 '20 at 16:31

score 2 · Answer 12 · edited Mar 23 '16 at 18:42

2

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

You can also use NSNumberFormatter class:

edited Mar 23 '16 at 18:42

hallucinations

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answered Jan 29 '14 at 10:33

user1479883

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score 2 · Answer 13 · answered Jul 07 '17 at 06:33

2

you can also do like this code 8.3.3 ios 10.3 support

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

answered Jul 07 '17 at 06:33

Parv Bhasker

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score 1 · Answer 14 · answered Aug 08 '13 at 07:18

1

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

answered Aug 08 '13 at 07:18

Sandip Patel - SM

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score 1 · Answer 15 · edited Nov 09 '17 at 08:10

1

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.

edited Nov 09 '17 at 08:10

Maple

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answered Nov 02 '17 at 06:23

Rizwan Ahmed

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score 0 · Answer 16 · answered Sep 23 '16 at 08:18

0

Worked in Swift 3

NSDecimalNumber(string: "Your string")

answered Sep 23 '16 at 08:18

user431791

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score 0 · Answer 17 · answered May 27 '17 at 07:27

0

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

answered May 27 '17 at 07:27

Ashu

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score -1 · Answer 18 · answered Oct 12 '17 at 18:55

-1

extension String {

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

let someFloat = "12.34".numberValue

answered Oct 12 '17 at 18:55

garg

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How to convert an NSString into an NSNumber

18 Answers18

Objective-C

Swift

Linked

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