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.
Related
I have a float var1 = cashInHandAmount = 4.73000002
I simply want as:
var2 = 4.73.
I have tried like this:
NSString *floatString = [NSString stringWithFormat:#"%.02f",cashInHandAmount];//it prints 4.73
[self.editcash setMaxValue:[floatString floatValue]];//but it again sets 4.73000002 why?
can you guys please help me regarding this?
%f simply rounds for the output. Transforming a float value into a string and back does not work, if the exact value (i. e. 4.73) has no representation in the float format. So transforming it back will "round" the stored value 4.73 into the float format, which is obviously 4.730…02.
You should rarely use (binary, IEEE) floats for financial calculating. Financial values (amount of money) is in most cases an integral value of cents, but no float value of dollars (or whatever). Additionally you can think about using NSDecimal and NSDecimalNumber to ensure, that every value with two digits of precision is storable in the format.
Edit:
float f = 4.73000002;
float rounded = roundf(f * 100.0f) / 100.0f;
NSLog(#"%10.10f", rounded);
outputs:
2014-07-01 10:25:48.653 xctest[596:303] 4.7300000191
It is difficult to check, but probably float can not represent 4.73 exactly. The nearest representable value is 4.7300000191. This is what I said: A rounded decimal representation is not always a possible "binary float" representation. You will face that problem with many values.
Try this :
float var1 = 4.73000002;
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
formatter.numberStyle = NSNumberFormatterDecimalStyle;
formatter.maximumFractionDigits = 2; //Set number of fractional digits
NSNumber *num = [NSNumber numberWithFloat:var1];
NSString *roundedNum = [formatter stringFromNumber:num];
DLOG(#"Answer : %#", roundedNum);
If you want truncate all other decimals you can declare a simple method like this:
- (float)sanitizeFloat:(float)value
{
float newValue = [[NSString stringWithFormat:#"%.2f",value] floatValue];
return newValue;
}
and then
float aFloat = 45.070809;
NSLog(#"%f",[self sanitizeFloat:aFloat]);
In your case:
[self.editcash setMaxValue:[self sanitizeFloat:cashInHandAmount]];
The output will be 45.070000
If you're working with monetary amounts that have fixed decimal places, you should always use NSDecimal or NSDecimalNumber - you shouldn't use floats at all, as they are inexact.
You could use:
-[NSDecimalNumber decimalNumberWithString:]
to create your number, and there are many methods for working with NSDecimalNumbers - see Apple's docs or NSDecimalNumber.h for details.
I have 53.025709438664 in string when I do [string_variable doubleValue], I get 53.025709.
I use NSLog(#"%f") to print it.
How I can get full precision value ?
Please have a look at the format specifiers provided in Apple doc.
You cannot get all the digits after decimal point.
Edit:
If you know exact number of digits after decimal, then we could use something as follows
NSString *s = #" 53.025709438664";
double d = [s doubleValue];
NSLog(#"%.12f",d);
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
Here is code:
NSNumber* number = [NSDecimalNumber decimalNumberWithString :#"11111111111111111"];
NSLog(#"%# = %lld", number, [number unsignedLongLongValue]);
And I get output:
11111111111111111 = 11111111111111112
Is it known bug or I do something wrong?
UPDATE:
bug reported: https://bugreport.apple.com/cgi-bin/WebObjects/RadarWeb.woa/3/wo/Stnn5Aaocknzx52c64zbN0/5.83.28.0.9
I believe the issue is that you are using NSNumber's -longLongValue method on an NSDecimalNumber. NSDecimalNumber stores a high-precision representation of the number internally as a decimal, but NSNumber just keeps it as an IEEE 754 double precision floating-point value. If I recall correctly, if you use the standard NSNumber superclass methods on an NSDecimalNumber, the internal number is first converted to a double, which can introduce binary floating-point artifacts like this.
As neoneye suggests, a way around this might be to use an NSString as an intermediary. For example:
NSDecimalNumber* number = [NSDecimalNumber decimalNumberWithString :#"11111111111111111"];
double number2 = [#"11111111111111111" doubleValue];
NSLog(#"%# = %lld = %f", number, [[number descriptionWithLocale:nil] longLongValue], number2);
Will produce the result 11111111111111111 = 11111111111111111 = 11111111111111112.000000, showing both the correct result from the decimal number extraction and the error that happens when this number is temporarily stored in a double.
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.