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.
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];
}
In my code, I'm dealing with an NSString that contains an NSNumber value. This NSNumber value could possibly be a repeating decimal number (e.x. 2.333333333e+06) that shortens to "2.333333" in a string format. It could also be a terminating number (e.x. 2.5), negative, or irrational number (2.398571892858...) (only dealing with decimals here)
I need to have a way to figure out if there are the repeating numbers in the string (or the NSNumber, if necessary). In my code, I would have no way to know what the repeating number would be, as it's a result of computations started by the user. I have tried this for loop (see below) that doesn't work the way I want it to, due to my inexperience with string indexing/ranges/lengths.
BOOL repeat = NO; //bool to check if repeating #
double repNum, tempNum; //run in for loop
NSString *repeating = [numVal stringValue]; //string that holds possible repeating #
for (int i = 3; i <= [repeating length]-3; i++) { //not sure about index/length here
if (i == 3) {
repNum = [repeating characterAtIndex:i];
}
tempNum = [repeating characterAtIndex:i];
if (tempNum == repNum) {
repeat = YES;
} else {
repeat = NO;
}
}
This code doesn't work as I'd like it to, mainly because I also have to account for negative dashes in the string and different amounts of numbers (13 1/3 vs. 1 1/3). I've used the modffunction to separate the integers from the decimals, but that hasn't worked well for me either.
Thank you in advance. Please let me know if I can clarify anything.
EDIT:
This code works with the finding of different solutions for polynomials (quadratic formula). Hope this helps put it into context. See here. (Example input)
NSNumber *firstPlusSolution, *secondMinusSolution;
NSString *pValueStr, *mValueStr;
firstPlusSolution = -(b) + sqrt(square(b) - (4)*(a)*(c)); //a, b, c: "user" provided
firstPlusSolution /= 2*(a);
secondMinusSolution = -(b) - sqrt(square(b) - 4*(a)*(c));
secondMinusSolution /= 2*(a);
pValueStr = [firstPlusSolution stringValue];
mValueStr = [secondMinusSolution stringValue];
if ([NSString doesString:pValueStr containCharacter:'.']) { //category method I implemented
double fractionPart, integerPart;
fractionPart = modf(firstPlusSolution, &integerPart);
NSString *repeating = [NSString stringWithFormat:#"%g", fractionPart];
int repNum, tempNum;
BOOL repeat = NO;
//do for loop and check for negatives, integers, etc.
}
if ([NSString doesString:mValueStr containCharacter'.']) {
//do above code
//do for loop and check again
}
Use C. Take the fractional part. Convert to a string with a known accuracy. If length of string indicates that last digits are missing, then it does not repeat. Use NSString-UTF8String to convert a string. Get rid of the last digit (may be rounding or actual floating point arithmetic error). Use function int strncmp ( const char * str1, const char * str2, size_t num ) to perform comparison within the string itself. If the result is 8 characters long and the last 2 characters match the first 2 characters, then shall the first 6 characters be considered repeating?
Assuming that fraction knowledge your desire:
• Possibility 1: Use fractions. Input fractions. Compute with fractions. Output fractions. Expand upon one of the many examples of a c++ fraction class if necessary and use it.
• Possibility 2: Choose an accuracy which is much less than double. Make a fraction from the result. Reduce the fraction allowing rounding based upon accuracy.
I suggest use not optimal but easy to write solution
Create NSMutableDictionary that will contain number as key and count of occurrence as value.
You can use componentsSeparatedByString: if numbers in string delimited by known symbol
In loop check valueForKey in dictionary and if need increase value
Last step is analyzing our dictionary and do anything you need with numbers
I'm trying to implement simple calculator using Objective-C. Here's my problem:
_currentNumber = #"4.65";
_num1 = [_currentNum intValue];
Obviously, this code above returns "4" instead of 4.65.
How could I make it return a decimal number?
The intValue will return an integer which is not a floating point number.
Use doubleValue or floatValue ,depending on the precession you need, to retrieve a floating point number.
You can try like this..
NSString *str=#"14.6";
CGFloat value=[str floatValue];
NSLog(#"%.2f",value);
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
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]