Convert a string into an int - ios

I'm having trouble converting a string into an integer. I googled it but all I can find is how to convert an int into a string. Does anyone know how to do it the other way around? Thanks.

See the NSString Class Reference.
NSString *string = #"5";
int value = [string intValue];

How about
[#"7" intValue];
Additionally if you want an NSNumber you could do
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter numberFromString:#"7"];

I use:
NSInteger stringToInt(NSString *string) {
return [string integerValue];
}
And vice versa:
NSString* intToString(NSInteger integer) {
return [NSString stringWithFormat:#"%d", integer];
}

This is the simple solution for converting string to int
NSString *strNum = #"10";
int num = [strNum intValue];
but when you are getting value from the textfield then,
int num = [txtField.text intValue];
where txtField is an outlet of UITextField

Swift 3.0:
Int("5")
or
let stringToConvert = "5"
Int(stringToConvert)

I had to do something like this but wanted to use a getter/setter for mine. In particular I wanted to return a long from a textfield. The other answers all worked well also, I just ended up adapting mine a little as my school project evolved.
long ms = [self.textfield.text longLongValue];
return ms;

NSString *string = /* Assume this exists. */;
int value = [string intValue];

Very easy..
int (name of integer) = [(name of string, no ()) intValue];

Yet another way: if you are working with a C string, e.g. const char *, C native atoi() is more convenient.

You can also use like :
NSInteger getVal = [self.string integerValue];

To convert an String number to an Int, you should do this:
let stringNumber = "5"
let number = Int(stringNumber)

Related

convert NSString to long value [duplicate]

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).
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

Assign the value from a text field into an int variable to do some math with other variables and then return it?

The code showing here assigns the value of an variables to the text field I want to do the opposite of that.
_Initial.text = [NSString stringWithFormat:#"%f", init];
The opposite of the code above. Please help
The opposite of that would be getting converting string to float. You can do it with this
init = [_Initial.text floatValue];
You should search for an answer before you ask a question.
NSString *intString = #"123";
NSString *floatString = #"456.789";
int intVal = [intString intValue];
float floatVal = [floatString floatValue];
printf("%i %f\n",intVal,floatVal);

Change NSString to int value?

I am having trouble trying to run this test due to an error on the first line of this code that states:
Incompatible pointer to integer conversion initializing 'int' with an
expression of type 'NSString *'
- (IBAction)Button:(id)sender
{
int x = TramNumber.text;
if (x < 9)
{
Tramresult.text = [NSString stringWithFormat:#"lol"];
}
else
{
NSLog (#"x is less than 9!");
}
}
#end
Please help. I am on iOS and running xCode 5.1.1 if that helps.
You are representing NSString value wrong.
Use this code sample to solve your problem:
int x = [TramNumber.text intValue];
To represent int value from your textfield.
int x = [TramNumber.text intValue];
You just need to convert NSString to int.
You cannot assign text as an integer.
You should cast NSString to int like this:
int x = [TramNumber.text intValue];
Just try this:
NSString *code = dic[#"code"];
int statusCode = [code intValue];
NSLog(#"%d",statusCode);
if(statusCode==1)//........cont...

Change numbers to text iOS [duplicate]

This question already has answers here:
How do I convert an integer to the corresponding words in objective-c?
(7 answers)
Closed 8 years ago.
I am not sure how to explain this but I need something like this and I'm not quite sure how to do it. I have a textfield where the user enters a number. For example "200". I got a label that must show "Two Hundred"(The number entered in Words )
Any ideas on how to do this?
Thanks in advance sorry for my bad english
try this:
//textField.text is 200
NSInteger someInt = [textField.text integerValue];
NSString *numberWord;
NSNumber *numberValue = [NSNumber numberWithInt:someInt];
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:NSNumberFormatterSpellOutStyle];
numberWord = [numberFormatter stringFromNumber:numberValue];
NSLog(#"numberWord= %#", numberWord); // Answer: two hundred
yourLabel.text = numberWord;
You can use a NSNumberFormatter It can convert an NSNumber into its word representation.
NSNumber* number = #100;
NSString* textNumber;
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:NSNumberFormatterSpellOutStyle];
textNumber = [numberFormatter stringFromNumber:number];
Here's something totally different. A bit more complex and weird, but you can fiddle around with it if you seek more manual control over the output. This is set up to serve a value up to 100,000 (to demonstrate a conditional set up).
Have fun:
int theInteger = 1,123;
int convert1000 = 0;
int convert100 = 0;
int convert10 = 0;
int convert1 = 0;
NSString * wordThousand = #"Thousand";
NSString * wordHundred = #"Hundred";
NSArray *wordsArraySingleDigit = [[NSArray alloc] initWithObjects:#"Zero",#"One",#"Two",#"Three",#"Four",#"Five",#"Six",#"Seven",#"Eight",#"Nine",nil];
NSArray *wordsArrayDoubleDigit = [[NSArray alloc] initWithObjects:#"Ten",#"Eleven",#"Twelve",...,#"Ninety Eight", #"Ninety Nine",nil];
convert1000 = (theInteger - (theInteger % 1000))/1000;
convert100 = ((theInteger - (convert1000 * 1000)) - ((theInteger - (convert1000 * 1000)) % 100))/100;
convert10 = ((theInteger - (convert1000 *1000)-(convert100 *100)) - ((theInteger -(convert1000 *1000) - (convert100 *100)) % 10))/10;
convert1 = (theInteger - (convert1000 *1000)-(convert100 *100) - (convert10 *10));
if (theInteger > = 10000 && < 100000){
NSString * convertedThousand = [wordsArrayDoubleDigit objectAtIndex : convert1000];
convertedThousand = [NSString stringWithFormat: #“%# %#“,convertedThousand,wordThousand];
NSLog (convertedThousand);
}
if (theInteger >= 1000 && <= 10000){
NSString * convertedThousand = [wordsArraySingleDigit objectAtIndex : convert1000];
convertedThousand = [NSString stringWithFormat: #“%# %#“,convertedThousand,wordThousand];
NSLog (convertedThousand);
}
NSString * convertedHundred = [wordsArraySingleDigit objectAtIndex : convert100];
convertedHundred = [NSString stringWithFormat: #“%# %#“, convertedHundred,wordHundred];
NSLog (convertedHundred);
NSString * convertedTen = [wordsArraySingleDigit objectAtIndex : convert10];
convertedTen = [NSString stringWithFormat: #“%#“, convertedTen];
NSLog (convertedTen);
NSString *convertedOne = [wordsArraySingleDigit objectAtIndex : convert1];
NSLog (convertedOne);
Hope this helps or gives you a starting point for an alternative approach.
I think you should account for every possibility and make a set of IF statements, for example, if the first digit of the number is 3, write "three", if the number has 4 digits, display "thousands"... of course you can organize your code to make the process easy.

How to do primary maths with cocoa and cocoa touch

I'm quite new to xcode and am making a app where you put two numbers in to a text input and I don't know how to make xcode to do the adding sum, I tried this but it did not work
self.answer.text = self.label.text + self.label2.text
Does anybody know how to do this.
Use :
NSInteger firstNumber=[self.label.text integerValue];
NSInteger secondNumber=[self.label2.text integerValue];
NSInteger total=firstNumber + secondNumber;
NSString *string=[NSString stringWithFormat:#"%d",total];
self.answer.text = string;
If you want to take double value replace integerValue with doubleValue
Here is another interesting way:
//Some string with expression which was taken from self.label.text
NSString *s = #"2*(2.15-1)-4.1";
NSExpression *expression = [NSExpression expressionWithFormat:s];
float result = [[expression expressionValueWithObject:nil context:nil] floatValue];
NSLog(#"%f", result);
self.answer.text=[NSString stringWithFormat:#"%f",result];

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