I need help with my calculator app.I'm using two float variables so when I do an exercise account, for example 6/2 the result is 3.000000 (with decimal point) and not 3 because it is a float variable. However, if I do 2/6 the result is 0.33333,How can I fix the problem that show the decimal point only if have a remainder.
Here's the code:
//Add the number to UILabel
-(IBAction)numberPressed:(UIButton *)sender {
NSString *number = sender.currentTitle;
if (self.typingNumber) {
self.firstLabel.text = [self.firstLabel.text stringByAppendingString:number];
}
else {
self.firstLabel.text = number;
self.typingNumber = YES;
}
}
//Equal method
-(IBAction)equalPress:(id)sender {
self.typingNumber = NO;
self.secondNum = [self.firstLabel.text floatValue];
result = 0;
if ([self.operation isEqualToString:#"+"]) {
result = self.firstNum + self.secondNum;
}
if ([self.operation isEqualToString:#"-"]) {
result = self.firstNum - self.secondNum;
}
if ([self.operation isEqualToString:#"X"]) {
result = self.firstNum * self.secondNum;
}
if ([self.operation isEqualToString:#"/"]) {
result = self.firstNum / self.secondNum;
}
self.firstLabel.text = [NSString stringWithFormat:#"%f" ,result];
}
You can try to use the %g format specification:
The double argument shall be converted in the style f or e (or in the style F or E in the case of a G conversion specifier), with the precision specifying the number of significant digits. If an explicit precision is zero, it shall be taken as 1. The style used depends on the value converted; style e (or E ) shall be used only if the exponent resulting from such a conversion is less than -4 or greater than or equal to the precision. Trailing zeros shall be removed from the fractional portion of the result; a radix character shall appear only if it is followed by a digit or a '#' flag is present.
from IEEE printf specification as used by Apple, emphasis added.
For example:
float a = 6.0 / 2.0;
float b = 2.0 / 6.0;
NSLog(#"%f, %g | %f, %g", a, a, b, b);
produces:
3.000000, 3 | 0.333333, 0.333333
For even more control use the NSNumberFormatter class, in particular look at properties such as alwaysShowsDecimalSeparator.
For example:
NSNumberFormatter *formatter = [NSNumberFormatter new];
formatter.numberStyle = NSNumberFormatterDecimalStyle;
formatter.alwaysShowsDecimalSeparator = NO;
NSLog(#"%# | %#", [formatter stringFromNumber:#(a)], [formatter stringFromNumber:#(b)]);
produces:
3 | 0.333
(you can use other properties to control the number of decimal places etc.)
Related
I want to round a double to one decimal place in Objective-C.
In Swift I can do it with an extension:
public extension Double {
/// Rounds the double to decimal places value
func rounded(toPlaces places:Int) -> Double {
let divisor = pow(10.0, Double(places))
return (self * divisor).rounded() / divisor
}
}
However, apparently you cannot call extensions on primitives from Objective-C so I can't use the extension.
I would be happy to do the rounding either on the double directly or as a string, however, neither of the following is working:
double mydub = 122.12022222223322;
NSString *axtstr = [NSString stringWithFormat:#"%2f", mydub]; //gives 122.120222
double rounded = (round(mydub*10)) / 10.0; //gives 122.100000
How do I convert 122.12022222223322; into 122.1?
You need to put a decimal between the % and 2f
[NSString stringWithFormat:#"%.2f", mydub];
double mydouble = 122.12022222223322;
NSString *str = [NSString stringWithFormat:#"%.2f", mydouble];
// = #"122.12"
.. will not round mydouble. Instead it will only apply format to the output as string.
double d = 122.49062222223322;
NSString *dStr = [NSString stringWithFormat:#"%.f %.1f %.2f %.3f", d, d, d, d];
// = #"122 122.5 122.49 122.491"
As Objective-C shares the language rules from C you can round safely with
#include <math.h>
double rounded = round(mydouble);
// = 122.000000
of course you can shift comma with multiplication and dividing the power of ten you want.
double commashifted = round(mydouble*100.0)/100.0;
// = 122.120000;
If you are really into Objective-C Classes to do same in deluxe have a look into 'NSDecimal.h' in the Foundation Framework.
Last but not least you can do the same with C as you did with swift.
double roundbycomma(int commata, double zahl) {
double divisor = pow(10.0, commata);
return round(zahl * divisor) / divisor;
}
So I've coded a calculator in objective C and I've formatted the display in .4g as I wanted only significant digits and 4 decimal places. This works fine :)
What I would like it to do is not display 2.34E+05 etc when it displays a longer number like 234,000. I've allowed it to autosize the text in the label so I know it isn't just that the label is too small.
Is there a piece of code that will make it display the actual number instead of the scientific notation?
Formatting with %f instead of %g won't use standard form.
Have a look at this specification.
Edit 1:
I found this answer to help with the rounding.
-(float) round:(float)num toSignificantFigures:(int)n {
if(num == 0) {
return 0;
}
double d = ceil(log10(num < 0 ? -num: num));
int power = n - (int) d;
double magnitude = pow(10, power);
long shifted = round(num*magnitude);
return shifted/magnitude;
}
I then combined these two options to get:
NSLog(#"%.0f", [self round:number toSignificantFigures:4]);
Edit 2:
What about this:
- (NSString *) floatToString:(float) val {
NSString *ret = [NSString stringWithFormat:#"%.5f", val];
unichar c = [ret characterAtIndex:[ret length] - 1];
while (c == 48 || c == 46) { // 0 or .
ret = [ret substringToIndex:[ret length] - 1];
c = [ret characterAtIndex:[ret length] - 1];
}
return ret;
}
I haven't tested this, but it looks as if it limits decimals to 5, then removes any trailing zeros or decimal points.
[Source]
I am trying to trim zeros after a decimal point as below but it's not giving desired result.
trig = [currentVal doubleValue];
trig = trig/100;
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
[formatter setMaximumFractionDigits:0];
display.text = [formatter stringFromNumber:[NSNumber numberWithDouble:trig]];
The number is still being displayed without trimming zeros after the decimal point.
Here currentVal is the number I am entering.
For example if i pass "trig" = 123 (Initially "trig" = 123 after doing trig/100 i want to display 1.23 but it is displaying as 1.23000000).
Sometimes the straight C format specifiers do an easier job than the Cocoa formatter classes, and they can be used in the format string for the normal stringWithFormat: message to NSString.
If your requirement is to not show any trailing zeroes, then the "g" format specifier does the job:
float y = 1234.56789f;
NSString *s = [NSString stringWithFormat:#"%g", y];
Notice that there is no precision information, which means that the printf library will remove the trailing zeroes itself.
There is more information in the docs, which refer to IEEE's docs.
In case this helps someone. I wanted 1 decimal value but no '.0' on the end if the float was '1.0'. Using %g would give scientific notation for longer numbers, following ugliness worked well enough for me as high accuracy wasn't critical.
// Convert to 1 dp string,
NSString* dirtyString = [NSString stringWithFormat: #"%.1f", self.myFloat];
// Convert back to float that is now a maximum of 1 dp,
float myDirtyFloat = [dirtyString floatValue];
// Output the float subtracting the zeros the previous step attached
return [NSString stringWithFormat: #"%g", myDirtyFloat];
This will not display any decimal value after the decimal point:
display.text = [NSString stringWithFormat:#"%1.0f", trig];
This will just trim the zeros after the decimal point:
isplay.text = [NSString stringWithFormat:#"%3.2f", trig];
display.text = [display.text stringByTrimmingCharactersInSet:[NSCharacterSet characterSetWithCharactersInString:[NSString stringWithFormat#"0"]]];
Note, this may leave you with the trailing decimal point. "124." may happen. I guess that some smarter solution will be posted soon.
From the documentation, it looks like setFractionDigits: is only for converting the other way.
The best thing to do is probably to convert your number to an integer before formatting it e.g.
double converted = round(trig); // man round for docs
You can use also the formatting functions of stringWithFormat: of NSString, but then you will lose all the localisation advantages you get with NSNumberFormatter.
This may not be a proper solution where there is NSNumberFormetter Class, But I just did this rather then googling a lot! ;)
Here is an example, if it helps:
-(NSString*) trimZerosAfterDecimalPoint:(NSString*)string_ {
double doubleValue=[string_ doubleValue];
long leftPart=(long)doubleValue;
double rightPart=doubleValue-(double)leftPart;
NSString *rightPartAsStr=[NSString stringWithFormat:#"%f", rightPart];
int i=0;
for (i=rightPartAsStr.length-1; i>=2; i--) {
if ([rightPartAsStr characterAtIndex:i]!='0') {
rightPartAsStr=[rightPartAsStr substringWithRange:NSMakeRange(2, i-1)];
break;
}
}
if (i<2) {
string_=[NSString stringWithFormat:#"%ld", leftPart];
} else {
string_=[NSString stringWithFormat:#"%ld.%#", leftPart, rightPartAsStr];
}
return string_;
}
I just had to do this for one of my programs and heres how I went about it:
- (void) simplify{
int length = (int)[self.calcString length];
for (int i = (int)[self.calcString length]; i > 0; i--) {
if ([self.calcString rangeOfString:#"."].location != NSNotFound) {
NSRange prevChar = NSMakeRange(i-1, 1);
if ([[self.calcString substringWithRange:prevChar] isEqualToString:#"0"]||
[[self.calcString substringWithRange:prevChar] isEqualToString:#"."])
length--;
else
break;
}
self.calcString = [self.calcString substringToIndex:length];
}
}
This works
display.text = [#(trig) stringValue];
it is because of your datatype cannot be formatted is such a manner.
I have the value 25.00 in a float, but when I print it on screen it is 25.0000000.
How can I display the value with only two decimal places?
It is not a matter of how the number is stored, it is a matter of how you are displaying it. When converting it to a string you must round to the desired precision, which in your case is two decimal places.
E.g.:
NSString* formattedNumber = [NSString stringWithFormat:#"%.02f", myFloat];
%.02f tells the formatter that you will be formatting a float (%f) and, that should be rounded to two places, and should be padded with 0s.
E.g.:
%f = 25.000000
%.f = 25
%.02f = 25.00
Here are few corrections-
//for 3145.559706
Swift 3
let num: CGFloat = 3145.559706
print(String(format: "%f", num)) = 3145.559706
print(String(format: "%.f", num)) = 3145
print(String(format: "%.1f", num)) = 3145.6
print(String(format: "%.2f", num)) = 3145.56
print(String(format: "%.02f", num)) = 3145.56 // which is equal to #"%.2f"
print(String(format: "%.3f", num)) = 3145.560
print(String(format: "%.03f", num)) = 3145.560 // which is equal to #"%.3f"
Obj-C
#"%f" = 3145.559706
#"%.f" = 3146
#"%.1f" = 3145.6
#"%.2f" = 3145.56
#"%.02f" = 3145.56 // which is equal to #"%.2f"
#"%.3f" = 3145.560
#"%.03f" = 3145.560 // which is equal to #"%.3f"
and so on...
You can also try using NSNumberFormatter:
NSNumberFormatter* nf = [[[NSNumberFormatter alloc] init] autorelease];
nf.positiveFormat = #"0.##";
NSString* s = [nf stringFromNumber: [NSNumber numberWithFloat: myFloat]];
You may need to also set the negative format, but I think it's smart enough to figure it out.
I made a swift extension based on above answers
extension Float {
func round(decimalPlace:Int)->Float{
let format = NSString(format: "%%.%if", decimalPlace)
let string = NSString(format: format, self)
return Float(atof(string.UTF8String))
}
}
usage:
let floatOne:Float = 3.1415926
let floatTwo:Float = 3.1425934
print(floatOne.round(2) == floatTwo.round(2))
// should be true
In Swift Language, if you want to show you need to use it in this way. To assign double value in UITextView, for example:
let result = 23.954893
resultTextView.text = NSString(format:"%.2f", result)
If you want to show in LOG like as objective-c does using NSLog(), then in Swift Language you can do this way:
println(NSString(format:"%.2f", result))
IN objective-c, if you are dealing with regular char arrays (instead of pointers to NSString) you could also use:
printf("%.02f", your_float_var);
OTOH, if what you want is to store that value on a char array you could use:
sprintf(your_char_ptr, "%.02f", your_float_var);
The problem with all the answers is that multiplying and then dividing results in precision issues because you used division. I learned this long ago from programming on a PDP8.
The way to resolve this is:
return roundf(number * 100) * .01;
Thus 15.6578 returns just 15.66 and not 15.6578999 or something unintended like that.
What level of precision you want is up to you. Just don't divide the product, multiply it by the decimal equivalent.
No funny String conversion required.
in objective -c is u want to display float value in 2 decimal number then pass argument indicating how many decimal points u want to display
e.g 0.02f will print 25.00
0.002f will print 25.000
Here's some methods to format dynamically according to a precision:
+ (NSNumber *)numberFromString:(NSString *)string
{
if (string.length) {
NSNumberFormatter * f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterDecimalStyle;
return [f numberFromString:string];
} else {
return nil;
}
}
+ (NSString *)stringByFormattingString:(NSString *)string toPrecision:(NSInteger)precision
{
NSNumber *numberValue = [self numberFromString:string];
if (numberValue) {
NSString *formatString = [NSString stringWithFormat:#"%%.%ldf", (long)precision];
return [NSString stringWithFormat:formatString, numberValue.floatValue];
} else {
/* return original string */
return string;
}
}
e.g.
[TSPAppDelegate stringByFormattingString:#"2.346324" toPrecision:4];
=> 2.3453
[TSPAppDelegate stringByFormattingString:#"2.346324" toPrecision:0];
=> 2
[TSPAppDelegate stringByFormattingString:#"2.346324" toPrecision:2];
=> 2.35 (round up)
Another method for Swift (without using NSString):
let percentage = 33.3333
let text = String.localizedStringWithFormat("%.02f %#", percentage, "%")
P.S. this solution is not working with CGFloat type only tested with Float & Double
Use NSNumberFormatter with maximumFractionDigits as below:
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
formatter.maximumFractionDigits = 2;
NSLog(#"%#", [formatter stringFromNumber:[NSNumber numberWithFloat:12.345]]);
And you will get 12.35
If you need to float value as well:
NSString* formattedNumber = [NSString stringWithFormat:#"%.02f", myFloat];
float floatTwoDecimalDigits = atof([formattedNumber UTF8String]);
lblMeter.text=[NSString stringWithFormat:#"%.02f",[[dic objectForKey:#"distance"] floatValue]];
Is there a simple way to do something like..
[NSMagicDataConverter humanStringWithBytes:20000000]
..which would return "19.1MB"?
Starting in OS X 10.8 and iOS 6, you can use NSByteCountFormatter.
Your example would look like this:
[NSByteCountFormatter stringFromByteCount:20000000 countStyle:NSByteCountFormatterCountStyleFile];
I would mush this into an NSFormatter subclass.
#import <Foundation/Foundation.h>
#interface SOFileSizeFormatter : NSNumberFormatter
{
#private
BOOL useBaseTenUnits;
}
/** Flag signaling whether to calculate file size in binary units (1024) or base ten units (1000). Default is binary units. */
#property (nonatomic, readwrite, assign, getter=isUsingBaseTenUnits) BOOL useBaseTenUnits;
#end
static const char sUnits[] = { '\0', 'K', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y' };
static int sMaxUnits = sizeof sUnits - 1;
#implementation SOFileSizeFormatter
#synthesize useBaseTenUnits;
- (NSString *) stringFromNumber:(NSNumber *)number
{
int multiplier = useBaseTenUnits ? 1000 : 1024;
int exponent = 0;
double bytes = [number doubleValue];
while ((bytes >= multiplier) && (exponent < sMaxUnits)) {
bytes /= multiplier;
exponent++;
}
return [NSString stringWithFormat:#"%# %cB", [super stringFromNumber: [NSNumber numberWithDouble: bytes]], sUnits[exponent]];
}
#end
Usage:
NSString *path = ...; // path to a file of 1,500,000 bytes
NSString *sizeString = nil;
NSNumber *sizeAttrib = [[[NSFileManager defaultManager] attributesOfItemAtPath:path error:NULL]objectForKey:NSFileSize];
SOFileSizeFormatter *sizeFormatter = [[[SOFileSizeFormatter alloc] init] autorelease];
[sizeFormatter setMaximumFractionDigits:2];
sizeString = [sizeFormatter stringFromNumber:sizeAttrib];
// sizeString ==> #"1.43 MB"
[sizeFormatter setUseBaseTenUnits:YES];
sizeString = [sizeFormatter stringFromNumber:sizeAttrib];
// sizeString ==> #"1.5 MB"
Here's my own take on the problem:
enum {
kUnitStringBinaryUnits = 1 << 0,
kUnitStringOSNativeUnits = 1 << 1,
kUnitStringLocalizedFormat = 1 << 2
};
NSString* unitStringFromBytes(double bytes, uint8_t flags){
static const char units[] = { '\0', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y' };
static int maxUnits = sizeof units - 1;
int multiplier = (flags & kUnitStringOSNativeUnits && !leopardOrGreater() || flags & kUnitStringBinaryUnits) ? 1024 : 1000;
int exponent = 0;
while (bytes >= multiplier && exponent < maxUnits) {
bytes /= multiplier;
exponent++;
}
NSNumberFormatter* formatter = [[[NSNumberFormatter alloc] init] autorelease];
[formatter setMaximumFractionDigits:2];
if (flags & kUnitStringLocalizedFormat) {
[formatter setNumberStyle: NSNumberFormatterDecimalStyle];
}
// Beware of reusing this format string. -[NSString stringWithFormat] ignores \0, *printf does not.
return [NSString stringWithFormat:#"%# %cB", [formatter stringFromNumber: [NSNumber numberWithDouble: bytes]], units[exponent]];
}
By default (if 0 is passed for flags), it will output SI units (base ten). You can set kUnitStringBinaryUnits to select binary (base two) units suitable for memory, or kUnitStringOSNativeUnits to have the unit type selected automatically based on OS version (pre-Leopard gets base two, post-Leopard gets base ten). Setting kUnitStringLocalizedFormat formats the string based on the user's current locale. For example:
unitStringFromBytes(1073741824, 0); // → "1.07 GB"
unitStringFromBytes(1073741824, kUnitStringBinaryUnits); // → "1 GB"
unitStringFromBytes(1073741824, kUnitStringOSNativeUnits | kUnitStringLocalizedFormat); // → "1.07 GB" (In Mac OS 10.6)
unitStringFromBytes(12345678901234567890123456789, kUnitStringOSNativeUnits | kUnitStringLocalizedFormat); // → "12,345.68 YB" (In Mac OS 10.6, in the US)
unitStringFromBytes(12345678901234567890123456789, kUnitStringOSNativeUnits | kUnitStringLocalizedFormat); // → "12.345,68 YB" (In Mac OS 10.6, in Spain)
Here's the helper function required for OS-native units:
BOOL leopardOrGreater(){
static BOOL alreadyComputedOS = NO;
static BOOL leopardOrGreater = NO;
if (!alreadyComputedOS) {
SInt32 majorVersion, minorVersion;
Gestalt(gestaltSystemVersionMajor, &majorVersion);
Gestalt(gestaltSystemVersionMinor, &minorVersion);
leopardOrGreater = ((majorVersion == 10 && minorVersion >= 5) || majorVersion > 10);
alreadyComputedOS = YES;
}
return leopardOrGreater;
}
NSString *stringFromFileSize(NSInteger theSize)
{
/*
From http://snippets.dzone.com/posts/show/3038 with slight modification
*/
float floatSize = theSize;
if (theSize<1023)
return([NSString stringWithFormat:#"%i bytes",theSize]);
floatSize = floatSize / 1024;
if (floatSize<1023)
return([NSString stringWithFormat:#"%1.1f KB",floatSize]);
floatSize = floatSize / 1024;
if (floatSize<1023)
return([NSString stringWithFormat:#"%1.1f MB",floatSize]);
floatSize = floatSize / 1024;
return([NSString stringWithFormat:#"%1.1f GB",floatSize]);
}
Here is a more Objective C-like function (uses NSNumber, NSArray, NSStirng, etc...) for doing this conversion.
This is based on Sidnicious's answer, so a big thanks for the initial work done there. Also based on Wikipedia articles.
Use it generally like this: [HumanReadableDataSizeHelper humanReadableSizeFromBytes:[NSNumber numberWithDouble:doubleValue]].
But, it looks like you want SI units with a 1024 multiplier so you would use it like this: [HumanReadableDataSizeHelper humanReadableSizeFromBytes:[NSNumber numberWithDouble:doubleValue] useSiPrefixes:YES useSiMultiplier:NO]
The reason I default to binary prefixes (ki, Mi) is because those seem to be the most appropriate unit prefix set to use for sizes of data on a computer. What you requested was the SI unit prefixes but using a multiplier of 1024, technically incorrect. Though I will note that SI prefixes for multiples of 1024 is fairly common and binary prefixes are not well accepted (according to Wikipedia).
HumanReadableDataSizeHelper.h
#interface HumanReadableDataSizeHelper : NSObject
/**
#brief Produces a string containing the largest appropriate units and the new fractional value.
#param sizeInBytes The value to convert in bytes.
This function converts the bytes value to a value in the greatest units that produces a value >= 1 and returns the new value and units as a string.
The magnitude multiplier used is 1024 and the prefixes used are the binary prefixes (ki, Mi, ...).
*/
+ (NSString *)humanReadableSizeFromBytes:(NSNumber *)sizeInBytes;
/**
#brief Produces a string containing the largest appropriate units and the new fractional value.
#param sizeInBytes The value to convert in bytes.
#param useSiPrefixes Controls what prefix-set is used.
#param useSiMultiplier Controls what magnitude multiplier is used.
This function converts the bytes value to a value in the greatest units that produces a value >= 1 and returns the new value and units as a string.
When useSiPrefixes is true, the prefixes used are the SI unit prefixes (k, M, ...).
When useSiPrefixes is false, the prefixes used are the binary prefixes (ki, Mi, ...).
When useSiMultiplier is true, the magnitude multiplier used is 1000
When useSiMultiplier is false, the magnitude multiplier used is 1024.
*/
+ (NSString *)humanReadableSizeFromBytes:(NSNumber *)sizeInBytes useSiPrefixes:(BOOL)useSiPrefixes useSiMultiplier:(BOOL)useSiMultiplier;
#end
HumanReadableDataSizeHelper.m
#implementation HumanReadableDataSizeHelper
+ (NSString *)humanReadableSizeFromBytes:(NSNumber *)sizeInBytes
{
return [self humanReadableSizeFromBytes:sizeInBytes useSiPrefixes:NO useSiMultiplier:NO];
}
+ (NSString *)humanReadableSizeFromBytes:(NSNumber *)sizeInBytes useSiPrefixes:(BOOL)useSiPrefixes useSiMultiplier:(BOOL)useSiMultiplier
{
NSString *unitSymbol = #"B";
NSInteger multiplier;
NSArray *prefixes;
if (useSiPrefixes)
{
/* SI prefixes
http://en.wikipedia.org/wiki/Kilo-
kilobyte (kB) 10^3
megabyte (MB) 10^6
gigabyte (GB) 10^9
terabyte (TB) 10^12
petabyte (PB) 10^15
exabyte (EB) 10^18
zettabyte (ZB) 10^21
yottabyte (YB) 10^24
*/
prefixes = [NSArray arrayWithObjects: #"", #"k", #"M", #"G", #"T", #"P", #"E", #"Z", #"Y", nil];
}
else
{
/* Binary prefixes
http://en.wikipedia.org/wiki/Binary_prefix
kibibyte (KiB) 2^10 = 1.024 * 10^3
mebibyte (MiB) 2^20 ≈ 1.049 * 10^6
gibibyte (GiB) 2^30 ≈ 1.074 * 10^9
tebibyte (TiB) 2^40 ≈ 1.100 * 10^12
pebibyte (PiB) 2^50 ≈ 1.126 * 10^15
exbibyte (EiB) 2^60 ≈ 1.153 * 10^18
zebibyte (ZiB) 2^70 ≈ 1.181 * 10^21
yobibyte (YiB) 2^80 ≈ 1.209 * 10^24
*/
prefixes = [NSArray arrayWithObjects: #"", #"ki", #"Mi", #"Gi", #"Ti", #"Pi", #"Ei", #"Zi", #"Yi", nil];
}
if (useSiMultiplier)
{
multiplier = 1000;
}
else
{
multiplier = 1024;
}
NSInteger exponent = 0;
double size = [sizeInBytes doubleValue];
while ( (size >= multiplier) && (exponent < [prefixes count]) )
{
size /= multiplier;
exponent++;
}
NSNumberFormatter* formatter = [[[NSNumberFormatter alloc] init] autorelease];
[formatter setMaximumFractionDigits:2];
[formatter setNumberStyle:NSNumberFormatterDecimalStyle]; // Uses localized number formats.
NSString *sizeInUnits = [formatter stringFromNumber:[NSNumber numberWithDouble:size]];
return [NSString stringWithFormat:#"%# %#%#", sizeInUnits, [prefixes objectAtIndex:exponent], unitSymbol];
}
#end
You can use FormatterKit and its TTTUnitOfInformationFormatter class:
https://github.com/mattt/FormatterKit
It is also available through CocoaPods with:
pod 'FormatterKit', '~> 1.1.1'
- (id)transformedValue:(id)value
{
double convertedValue = [value doubleValue];
int multiplyFactor = 0;
NSArray *tokens = #[#"bytes",#"KB",#"MB",#"GB",#"TB"];
while (convertedValue > 1024) {
convertedValue /= 1024;
multiplyFactor++;
}
return [NSString stringWithFormat:#"%4.2f %#",convertedValue, tokens[multiplyFactor]];
}
I know the questions is for Obj C but if anyone looking for a swift version:
public static func fileSizeDisplay(fromBytes:Int) -> String {
let display = ["bytes","KB","MB","GB","TB","PB"]
var value:Double = Double(fromBytes)
var type = 0
while (value > 1024){
value /= 1024
type = type + 1
}
return "\(String(format:"%g", value)) \(display[type])"
}