To get a square root value of mainlabelString, I am using
- (IBAction)rootPressed:(id)sender
{
NSString *mainLabelString = mainLabel.text;
int mainLabelValue = [mainLabelString longLongValue];
NSString *calculatedValue = [NSString stringWithFormat: #"%f", sqrt(mainLabelValue)];
mainLabel.text = calculatedValue;
}
And although it does work with numbers such as 88, from which I get 9.380832, it does not for example work with that number, for which it says the square root is 3.000000 (instead of 3.062814).
I tried replacing longLongValue with doubleValue and integerValue but it doesn't change it.
What's wrong?
If the value into the text field is like #"9.0001", getting the long long value truncates the decimal part.You said that you also have tried doubleValue, but I suspect that you did something like this:
int mainLabelValue = [mainLabelString doubleValue];
Instead of:
double mainLabelValue = [mainLabelString doubleValue];
In the first case the number loses the decimal part too, because an int can't store the decimal part, nor a long long int.
Let me know how you exactly tried to retrieve the double value of the text field.
Use NSNumberFormatter it provides very wide range and different styles(Decimaland scientific etc).documentation
Related
I have this line of code
// valueX is a long double (long double is a huge floating point)
NSString *value = [NSString stringWithFormat: #"%.10Lg", valueX];
This format specifier is specifying up to 10 decimal digits but I don't want to hard code this to 10.
I have this variable numberOfDigits that I want to be used to define the number of digits. For those itching to down vote this question, it is not so easy as it seems. I cannot substitute the 10 with %# because %.10Lg is a format specifier by itself.
OK, I can create a bunch of strings like #"%.5Lg", #"%.8Lg", #"%.9Lg"... and switch that, but I wonder if there is another way...
There is, if you read the manual pages for format specifiers. You can replace the precision with *, which means it will get taken from a parameter instead.
int numDigits = 10;
NSString *value = [NSString stringWithFormat:#"%.*Lg", numDigits, valueX];
I couldn't find this in the core foundation reference, but I know that this is written in the man 3 printf man page.
Dietrich's answer is the simplest and therefore best. Note that even if there wasn't a built-in way to specify the number of digits with a parameter you could still have done it by first building your format string and then using it:
- (NSString *) stringFromValue: (long double) value digits: (int) digits; {
//First create a format string. Use "%%" to escape the % escape char.
NSString *formatString =[NSString stringWithFormat: #"%%.%dLg", digits];
return [NSString stringWithFormat: formatString, value];
}
I am trying to convert quite big double number to int (or long or long long), but have some difficulties. Almost always it converts good, but not sometimes:
My code:
double price = 12345678.900000;
double hundredNumber = price * 100;
NSNumber *number = [NSNumber numberWithDouble:hundredNumber];
int tempNumber = [number intValue];
All goes good, until tempNumber. it logs out 1234567889, but it should be 1234567890 (...89 - ...90)
Does anyone know why it could happen and how to convert correctly?
P. S. I am trying to implement backspace to value (e.x. 123.45, after that it should be 12.34). Maybe anyone had implemented something like this?
You're always going to get the risk of rounding errors if you're using floating point numbers.
Why not always store prices as a long long?
i.e. instead of £5.50, store 550p. That way you will never have any rounding issues at all.
As commented, I would be careful with the roundings, because of the possible errors.
One possible solution is to work with doubles like Google does with coordinates in Android: multiplying them by 1E6. If you operate with integers then you'll safe much more CPU cycle than operating with doubles. Try this out:
double priceDouble = 33.f / 34.f;
NSLog(#"double: %f", priceDouble);
NSInteger priceInteger = (NSInteger)(priceDouble * 1E6);
NSLog(#"int: %d", priceInteger);
NSNumber * priceNumberWithDouble = [NSNumber numberWithDouble:priceInteger];
priceDouble = [priceNumberWithDouble doubleValue];
NSLog(#"double: %f", priceDouble);
NSNumber * priceNumberWithInteger = [NSNumber numberWithInteger:priceInteger];
priceInteger = [priceNumberWithInteger integerValue];
NSLog(#"int: %d", priceInteger);
double test = ((double)priceInteger)/1E6;
NSLog(#"Test: %f",test);
My output is the following:
double: 0.970588
int: 970588
double: 970588.000000
int: 970588
Test: 0.970588
I'm trying to make a unit converter as my first app. I've been having trouble getting the numeric value from the textfield, so I can multiply it by whatever constant to convert the units. Also, how can I add a decimal place to a numeric key pad?
Use the NSString doubleValue property to convert the string value from the UITextBox into a double.
double d = [str doubleValue];
float value = [self.myTextField.text floatValue];
I need to change the code below to make "intAmount" a decimal or an integer (i.e. a person can enter .10 or 1) in my uitextfield. The last line "myProduct" has to be a decimal not an integer and return the product in the format "18.00" for example. Can someone help someone help me alter my code snippit for this?
//amt has to be converted into a decimal value its a NSString now
NSInteger intAmount = [amt intValue];
//where total gets updated in the code with some whole (integer) value
NSInteger total=0;
//Change myProduct to a decimal with presicion of 2 (i.e. 12.65)
NSInteger myProduct=total*intAmount;
THIS DOESN'T WORK
NSDecimalNumber intAmount = [amt doubleValue];
//Keep in mind totalCost is an NSInteger
NSDecimalNumber total=totalCost*intAmount;
Use doubleValue instead of intValue to get the correct fractional number out of your text field. Put it in a variable of type double rather than NSInteger. Then use the format %.2g when you print it out and it will look like you want it to.
If you need to track decimal values explicitly, you can use NSDecimalNumber. However, if all you're doing is this one operation, Carl's solution is most likely adequate.
If you have a string representation of a real number (non-integer), you can use an NSScanner object to scan it into a double or float, or even an NSDecimal structure if that is your true intention (the NSDecimal struct and NSDecimalNumber class are useful for containing numbers that can be exactly represented in decimal).
NSString *amt = #"1.04";
NSScanner *aScanner = [NSScanner localizedScannerWithString:amt];
double theValue;
if ([aScanner scanDouble:&theValue])
{
// theValue should equal 1.04 (or thereabouts)
}
else
{
// the string could not be successfully interpreted
}
The benefit to using a localised NSScanner object is that the number is interpreted based on the user's locale, because “1.000” could mean either one-thousand or just one, depending on your locale.
I want to convert a string into a double and after doing some math on it, convert it back to a string.
How do I do this in Objective-C?
Is there a way to round a double to the nearest integer too?
You can convert an NSString into a double with
double myDouble = [myString doubleValue];
Rounding to the nearest int can then be done as
int myInt = (int)(myDouble + (myDouble>0 ? 0.5 : -0.5))
I'm honestly not sure if there's a more streamlined way to convert back into a string than
NSString* myNewString = [NSString stringWithFormat:#"%d", myInt];
To really convert from a string to a number properly, you need to use an instance of NSNumberFormatter configured for the locale from which you're reading the string.
Different locales will format numbers differently. For example, in some parts of the world, COMMA is used as a decimal separator while in others it is PERIOD — and the thousands separator (when used) is reversed. Except when it's a space. Or not present at all.
It really depends on the provenance of the input. The safest thing to do is configure an NSNumberFormatter for the way your input is formatted and use -[NSFormatter numberFromString:] to get an NSNumber from it. If you want to handle conversion errors, you can use -[NSFormatter getObjectValue:forString:range:error:] instead.
Adding to olliej's answer, you can convert from an int back to a string with NSNumber's stringValue:
[[NSNumber numberWithInt:myInt] stringValue]
stringValue on an NSNumber invokes descriptionWithLocale:nil, giving you a localized string representation of value. I'm not sure if [NSString stringWithFormat:#"%d",myInt] will give you a properly localized reprsentation of myInt.
Here's a working sample of NSNumberFormatter reading localized number String (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 such as "8,765.4 ", 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
olliej's rounding method is wrong for negative numbers
2.4 rounded is 2 (olliej's method gets this right)
−2.4 rounded is −2 (olliej's method returns -1)
Here's an alternative
int myInt = (int)(myDouble + (myDouble>0 ? 0.5 : -0.5))
You could of course use a rounding function from math.h
// Converting String in to Double
double doubleValue = [yourString doubleValue];
// Converting Double in to String
NSString *yourString = [NSString stringWithFormat:#"%.20f", doubleValue];
// .20f takes the value up to 20 position after decimal
// Converting double to int
int intValue = (int) doubleValue;
or
int intValue = [yourString intValue];
For conversion from a number to a string, how about using the new literals syntax (XCode >= 4.4), its a little more compact.
int myInt = (int)round( [#"1.6" floatValue] );
NSString* myString = [#(myInt) description];
(Boxes it up as a NSNumber and converts to a string using the NSObjects' description method)
For rounding, you should probably use the C functions defined in math.h.
int roundedX = round(x);
Hold down Option and double click on round in Xcode and it will show you the man page with various functions for rounding different types.
This is the easiest way I know of:
float myFloat = 5.3;
NSInteger myInt = (NSInteger)myFloat;
from this example here, you can see the the conversions both ways:
NSString *str=#"5678901234567890";
long long verylong;
NSRange range;
range.length = 15;
range.location = 0;
[[NSScanner scannerWithString:[str substringWithRange:range]] scanLongLong:&verylong];
NSLog(#"long long value %lld",verylong);
convert text entered in textfield to integer
double mydouble=[_myTextfield.text doubleValue];
rounding to the nearest double
mydouble=(round(mydouble));
rounding to the nearest int(considering only positive values)
int myint=(int)(mydouble);
converting from double to string
myLabel.text=[NSString stringWithFormat:#"%f",mydouble];
or
NSString *mystring=[NSString stringWithFormat:#"%f",mydouble];
converting from int to string
myLabel.text=[NSString stringWithFormat:#"%d",myint];
or
NSString *mystring=[NSString stringWithFormat:#"%f",mydouble];
I ended up using this handy macro:
#define STRING(value) [#(value) stringValue]