I'm not sure what's going on here. For some reason NSString seems unwilling to load in the double value d.
Here's my code:
-(NSString*)minuteFormat:(double) d{
NSString* mystring;
if(d >= 10){
mystring = [NSString stringWithFormat:#"%d", d];
}else{
mystring = [NSString stringWithFormat:#"0%d",d];
}
return(mystring);
}
Regardless of the value of d, the only thing that's getting returned is 0 or 00. (And I'm sure d is getting inputted correctly, as I've used breakpoints to check.)
Could someone tell me what's going on?
%d is for integers. You probably want %f. See String Format Specifiers
%d is a decimal integer format string. Try %lf for double.
You should use %f (double) like this
mystring = [NSString stringWithFormat:#"%f", f];
or do a typecast of the value. For this you may use [yourString floatValue]
Related
I've been working on a calculator and I wanted to implement conversions from decimal to octal and from decimal to hexadecimal. I'm new to Xcode and Objective C, but I've managed to get a conversion from decimal to octal, it just doesn't seem to work with hexadecimal.
Here's the code I've written to convert a double to octal:
double result = 0;
...
double decToOct = [self popOperand];
NSString *oct = [NSString stringWithFormat:#"%llo", (long long)decToOct];
result = [oct doubleValue];
Using the same scheme (obviously that includes changing #"%llo" with #"%llx") the conversion to hexadecimal works up to a certain point. It does numbers 0 through 9 just fine, but once it hits 10, it comes up as 0. To test, I also input 5395 and it displayed 1513, the desired result.
Because of this, I can only assume that for some reason my code does not want to input the actual letters of the hexadecimal values (e.g. 11 would convert to B but it shows up as 0) .
Any suggestions? Thanks in advance.
UPDATE:
In addition, I have also been using this to display the result:
double result = [self.brain performOperation:operation];
self.display.text = [NSString stringWithFormat:#"%g", result];
result, as listed from the top, is an argument which is eventually returned here, to self.brain performOperation:operation. This is supposed to handle the display of all operations, including: addition, multiplication, etc. but also octal and hexadecimal. Again, it works fine with octal, but not with hexadecimal.
Try this, May be it will help you. Please do let me know if i am wrong here:--->
NSString *decStr = #"11";
NSString *hexStr = [NSString stringWithFormat:#"%lX",
(unsigned long)[dec integerValue]];
NSLog(#"%#", hexStr);
If you know your string only contains a valid decimal number then the simplest way would be:
NSString *dec = #"254";
NSString *hex = [NSString stringWithFormat:#"0x%lX",
(unsigned long)[dec integerValue]];
NSLog(#"%#", hex);
I'm struggling with concatenating NSStrings and ints. In isolation the below code works great. It returns "This is a test string with an int 10"
int myInt =10;
NSString *newstring =
[NSString stringWithFormat: #"This is a test string with an int %i", myInt];
NSLog(#"%#", newstring);
However when I put the below code into my project i get an error: Implicit conversion of int to NSString is disallowed with arc."
[_mycrop setTempLeft: (#"left value %i is %i", count, [_mycrop leftValue])];
Could anybody suggest where I'm going wrong? Although im passing in 2 variables, to my mind both are essentially the same.
The code "in isolation" is very different from the second code.
You have to use stringWithFormat: in the second example too.
[_mycrop setTempLeft:[NSString stringWithFormat:#"left value %i is %i", count, [_mycrop leftValue]]];
Or with two lines but easier to understand:
NSString *tempLeft = [NSString stringWithFormat:#"left value %i is %i", count, [_mycrop leftValue]];
[_mycrop setTempLeft:tempLeft];
Some documentation: Apple String Programming Guide - Formatting String Objects
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 got a problem with converting an NSNumber value to an NSString
MyPowerOnOrNot is an NSNumber witch can only return a 1 or 0
and myString is an NSString..
myString = [NSString stringWithFormat:#"%d", [myPowerOnOrNot stringValue]];
NSLog(#"%#",myString);
if(myString == #"1") {
[tablearrayPOWERSTATUS addObject:[NSString stringWithFormat:#"%#",#"ON"]];
}
else if(myString == #"0") {
[tablearrayPOWERSTATUS addObject:[NSString stringWithFormat:#"%#",#"OFF"]];
}
What is wrong with this?
The NSLog shows 0 or 1 in the console as a string but I can't check it if it is 1 or 0 in an if statement?
If doesn't jump into the statements when it actually should.. I really don't understand why this doesn't works..
Any help would be very nice!
A couple of problems
myString = [NSString stringWithFormat:#"%d", [myPowerOnOrNot stringValue]];
-stringValue sent to an NSNumber gives you a reference to a string. The format specifier %d is for the C int type. What would happen in this case is that myString would contain the address of the NSString returned by [myPowerOnOrNot stringValue]. Or, on 64 bit, it would return half of that address. You could actually use [myPowerOnOrNot stringValue] directly and avoid the relatively expensive -stringWithFormat:
if(myString == #"1")
myString and #"1" are not necessarily the same object. Your condition only checks that the references are identical. In general with Objective-C you should use -isEqual: for equality of objects, but as we know these are strings, you can use -isEqualToString:
if ([[myPowerOnOrNot stringValue] isEqualToString: #"1"])
Or even better, do a numeric comparison of your NSNumber converted to an int.
if ([myPowerOnOrNot intValue] == 1)
Finally if myPowerOnOrNot is not supposed to have any value other than 0 or 1, consider having a catchall else that asserts or throws an exception just in case myPowerOnOrNot accidentally gets set wrong by a bug.
"myString " is a reference to a string, not the value of the string itself.
The == operator will compare the reference to your string literal and so never return true.
Instead use
if( [myString isEqualToString:#"1"] )
This will compare the value of myString to "1"
In Objective C; you can't compare strings for equality using the == operator.
What you want to do here is as follows:
[tablearrayPOWERSTATUS addObject:([myPowerOnOrNot integerValue]?#"ON":#"OFF"])];
Compact, fast, delicious.
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]