I have strings that look about like this:
stringA = #"29.88";
stringB = #"2564";
stringC = #"12";
stringD = #"-2";
what is the best way to convert them so they can all be used in the same mathmatical formula?? that includes add, subtract.multiply,divide etc
Probably floatValue (as it appears you want floating-point values), though integerValue may also be of use (both are instance methods of NSString).
[stringA doubleValue]
These are all wrong, because they don't handle errors well. You really want an NSNumberFormatter.
If you have the string #"abc" and try to use intValue or floatValue on it, you'll get 0.0, which is obviously incorrect. If you parse it with an NSNumberFormatter, you'll get nil, which is very easy to distinguish from an NSNumber (which is what would be returned if it was able to parse a number).
Assuming that you have NSString variables.
NSString *stringA = #"29.88";
NSString *stringB = #"2564";
NSString *stringC = #"12";
NSString *stringD = #"-2";
suppose, you want to convert a string value to float value, use following statement.
float x=[stringA floatValue];
suppose, you want to convert a string value to integer value, use following statement.
NSInteger y = [stringC intValue];
Hope, it helps to you.
Related
I am making a calculator that logs input in a label named "inputLabel' and then outputs the answer in a different label named "outputLabel" (similar to a graphing calculator). Once the user is finished entering the expression, the expression is stored in an NSString object and then parsed with the NSPredicate class and evaluated with the NSExpression class. What I have works, but I have noticed for particular operations the answers are not correct. For example, if the user types in "25/2" the calculator returns 12, which is obviously incorrect. However, if the user types in "25/2.0" or "25.0/2" the calculator returns 12.5 which is what I want. It seems that the NSExpression method 'expressionValueWithObject' is interpreting the operands as integers instead of floats. If this is the case, is there a way that I change the 'expressionValueWithObject'method to interpret the operands as floats?
Brain.m
-(float)performCalculation: (NSString *)operation
{
NSPredicate *parsed = [NSPredicate predicateWithFormat:[operation stringByAppendingString:#"=1.0"]];
NSExpression *inputExpressionParsed = [(NSComparisonPredicate *)parsed leftExpression];
NSNumber *result = [inputExpressionParsed expressionValueWithObject:inputExpressionParsed context:nil];
return [result floatValue];
}
ViewController.m
- (IBAction)equalsPressed:(id)sender
{
//self.inputLabel.text = [self.inputLabel.text stringByAppendingString:#".0"];
NSString *inputExpression = self.inputLabel.text;
self.inputLabel.text = [self.inputLabel.text stringByAppendingString:#"="];
float result = [self.brain performCalculation:inputExpression];
self.outputLabel.text = [NSString stringWithFormat:#"%g", result];
}
No, NSExpression cannot do that. You could try to append ".0" to all integer numbers
in the string before evaluating it, but the better solution is probably to use a "proper"
math expression parser, for example
https://github.com/davedelong/DDMathParser
You could iterate through the expression tree replacing the expression with the integer value (expressionType == NSConstantExpression). It depends a little bit of the features of your calculator, whether it is worth or not.
I'm trying to create NSDecimalNumber with simply format like: 123.00 with two fractional digits after dot, always. I tried use the NSFormatter and many other ways like converting float from string and creating then NSDecimalNumber from this string, but it's not working.
The problem is that I need only NSDecimalNumber in this format, not NSString or any other.
Thanks for any advice,
Paul
You may get idea from this.
float num = 123.1254546;
NSString *str = [NSString stringWithFormat:#"%.2f",num];
NSLog(#"%.2f %#",num,str);
I think you simply need to do Type Casting operation for Two times as bellow
float value = 123.562;
int myDigit = value;
it gives value 123 in myDigit variable
NSLog(#"MyDigit in Decimal = %d",myDigit);
Output is MyDigit in Decimal = 123
now if you want output like 123.000 then simply write as bellow
float valueWithZiro = myDigit;
NSLog(#"MyDigit with 000 == %3f",valueWithZiro);
Output is MyDigit in Decimal = 123.000
NSDecimalNumber, like NSNumber, cannot contain formatting information. The object structure simply doesn't support it. It represents a number, not the way the number is displayed.
You can convert it to a formatted NSString (which you say you don't want). But you can't do what you're asking.
You convert it to a formatted NSString using an NSNumberFormatter. It's the object that allows you to specify the decimal and thousands separators, number of decimal places to display, the currency symbol, etc.
Maybe you were looking to store a NSNumberDecimal with just two digits after the fraction?
If so NSDecimalNumberBehaviors is your friend.
I had a similar need and used the following:
self.decimalHandlingBehaviorForApp = [NSDecimalNumberHandler
decimalNumberHandlerWithRoundingMode:NSRoundUp
scale:2 raiseOnExactness:NO
raiseOnOverflow:NO raiseOnUnderflow:NO
raiseOnDivideByZero:YES];
Edit: added example of using it
// update the taxable total first
self.cartTaxableTotal = [self.cartTaxableTotal decimalNumberByAdding:itemAdded.priceOfItem
withBehavior:self.decimalHandlingBehaviorForApp];
I need to convert NSDecimalNumber with lat/lon values which I need to convert to CLLocationDegrees. I used -[NSDecimalNumber doubleValue] method. But the value loses its precision. I want the values to be same. The following is what I am talking about(I hope everyone would be aware about this issue already).
NSString *coordStr = #"-33.890934125621094";
NSDecimalNumber *lat = [NSDecimalNumber decimalNumberWithString:coordStr];
NSLog(#"%#", lat); // -33.890934125621094
NSLog(#"%lf", [lat doubleValue]); // -33.890934
Is there any way that I should do instead of doing the above?
Try out to convert the String with the doubleValue method
NSString *coordStr = #"-33.890934125621094";
NSLog(#"%.17g",coordStr.doubleValue); //-33.890934125621094
Check out this post: How to print a double with full precision on iOS?
edit:
the docu says: typedef double CLLocationDegrees; so you can double test = coordStr.doubleValue; the problem is only, that NSLog doesn't print the complete value. but instead the double var saves the complete value.
CLLocationDegree is a double. So the maximum precision can only equal that of a double datatype.
How can I remove the all the characters after the decimal point.
Instead of 7.3456, I would just like 7.
This is what I do to get the number so far with decimal places.
[NSString stringWithFormat:#" %f : %f",(audioPlayer.currentTime),(audioPlayer.duration) ];
Many Thanks,
-Code
You can specify what you want using format string :
[NSString stringWithFormat:#" %.0f : %.0f", (audioPlayer.currentTime),
(audioPlayer.duration)];
If you want this for display, use an NSNumberFormatter:
double sevenpointthreefourfivesix = 7.3456;
NSNumberFormatter * formatter = [[NSNumberFormatter alloc] init];
[formatter setMaximumFractionDigits:0];
NSLog(#"%#", [formatter stringFromNumber:[NSNumber numberWithDouble:sevenpointthreefourfivesix]]);
2011-12-20 20:19:48.813 NoDecimal[55110:903] 7
If you want a value without the fractional part, use round(). If you want the closest integer value not greater than the original value, use floor().
floorf() is the function you're looking for.
you are after
[NSString stringWithFormat:#" %.00f : %.00f",(audioPlayer.currentTime),(audioPlayer.duration) ];
When formatting float you can tell the precision by the number before the f
Cast to int:
[NSString stringWithFormat:#" %i : %i",(int)(audioPlayer.currentTime),(int)(audioPlayer.duration) ];
Casting like this always rounds down (eg: just removes everything after the decimal place). This is what you asked for.
In the case of rounding to the NEAREST whole number you want to add 0.5 to the number
[NSString stringWithFormat:#" %i : %i",(int)(audioPlayer.currentTime+0.5f),(int)(audioPlayer.duration+0.5f) ];
This will round to the nearest whole number. eg: 1.2 becomes 1.7 and casting to int makes 1. 3.6 becomes 4.1 and casting makes 4. :)
Why not just cast the audioPlayer.currentTime to an integer before you use stringWithFormat?
[NSString stringWithFormat:#"%d", (int)(audioPlayer.currentTime)];
All you need to do is type-cast the double to an int, like so: int currentTime_int = (int)audioPlayer.currentTime;.
You can use this same approach for the other variable.
Many of the shorter answers here will work correctly. But if you want your code to be really clear and readable, you might want to explicitly specify your desired conversion from float to int, such as using:
int tmpInt = floorf(myFloat); // or roundf(), etc.
and then separately specifying how you want the integer formated, e.g.
... stringWithFormat:#"%d", tmpInt ... // or #"%+03d", etc.
instead of assuming that an inline cast shows what you want.
You may also use
double newDvalue =floor(dValue);
it will remove all the decimals point
using %.0f for string format will be good also
I'm trying to convert an NSString object to an NSNumber with the same numerical value. When I create the NSString object using this statement...
NSString *songID = [localNotif.userInfo objectForKey:#"SongID"];
the var songID contains the string value #"10359537395704663785". and when I attempt to convert it to an NSNumber object using the statement...
NSNumber *songIDNumber = [NSNumber numberWithLongLong:[songID longLongValue]];
the var songIDNumber contains the wrong value of 9223372036854775807
What am I doing wrong? It might also be worth noting that sometimes this code does work and produce the correct NSNumber value, but most of the time it fails as shown above.
Thanks in advance for your help!
UPDATE: God I love this site! Thanks to unbeli and carl, I was able to fix it using the updated code for converting from the NSString to the NSNumber...
unsigned long long ullvalue = strtoull([songID UTF8String], NULL, 0);
NSNumber *songIDNumber = [NSNumber numberWithUnsignedLongLong:ullvalue];
longLongValue returns LLONG_MAX in case of overflow, and LLONG_MAX is exactly what you get, 9223372036854775807. You value simply does not fit in long long
Try using NSDecimalNumber instead.
9223372036854775807 decimal is 0x7FFFFFFFFFFFFFFF in hex. Your number is 10359537395704663785, which is too large for a long long and overflows. From the NSString documentation:
Returns LLONG_MAX or LLONG_MIN on overflow.