For the code below:
double j1;
j1=7000000 //example
ItemE[5]=[NSString stringWithFormat:#"#1. total inc = %g", j1];
ItemE[5] is returned as "1.total inc = 7e +06"
How do I prevent the scientific notation and have "1.total inc = 7000000" instead?
Use %f:
ItemE[5]=[NSString stringWithFormat:#"#1. total inc = %f", j1];
Edit:
If you don't want decimal places you should use:
ItemE[5]=[NSString stringWithFormat:#"#1. total inc = %.f", j1];
To elaborate, you were using wrong specifier in format string.%g instructs to create string representation of floating-point variable in scientific notation. Normally you should use %f to represent double and float variable. By default, this specifier will result in number with 6 decimal places. In order to change that you can modify that specifier, for example:
%5.3f means that string should have 3 decimal places and should be 5 characters long. That means that if representation would be shorter than 5 chars, string will have additional spaces in front of number to give 5 chars total. Note that if you will have large number, it'll not be truncated. Consider code:
double pi = 3.14159265358979323846264338327950288;
NSLog(#"%f", pi);
NSLog(#"%.3f", pi);
NSLog(#"%8.3f", pi);
NSLog(#"%8f", pi);
NSLog(#"%.f", pi);
will give result:
3.141593
3.142
3.142
3.141593
3
please, try to use this one:
double j1 = 7000000.f;
NSLog(#"1. total inc = %.f", j1);
the result will be:
1. total inc = 7000000
I hope it helps.
Related
I am trying to convert an int to a string in objective-C.
I read the other questions on SO about converting ints to strings, and I tried this method in my code:
-(void)setCounter:(int)count
{
counterText.text = [NSString stringWithFormat:#"%d",count];
}
However, if I want to display a number like '01' the 0 is taken out of the conversion and only '1' is displayed. Is there a workaround?
There is no such number as 01. If you write
int count = 01;
it is compiled equivalently to
int count = 1;
In fact, be careful: 07 is equivalent to 7, but 011 is equivalent to 9!
What you can do is ask stringWithFormat: to give you the zero-padding:
[NSString stringWithFormat:#"%02d",count]
should give you "02" if count is 2. To deconstruct it:
% - interpolate the next value here
0 - pad it to the width by placing zeroes on the left side
2 - width is 2 characters
d - it will be an integer. Do it now.
If you want a different format to the one shown, use it:
counterText.text = [NSString stringWithFormat:#"%02d",count];
There are a huge range of possibilities with the format string.
if any number start from 0 then Its a octal representation (0 - 7). you can add zero explictly using below line.
counterText.text = [NSString stringWithFormat:#"0%d",count];
Check out..
NSNumber +numberWithInt
and then:
NSNumberFormatter -setMinimumIntegerDigits
and then get the string representation with:
NSString -stringFromNumber
how to display exponent value after calulation to textbox in iphone sdk.For example say 6.4516e-10.i am not getting answer for it in my textbox after calculating 10 * 6.4516e-10.please tell me solution..
Use StringWithFormat and exponent formations:
From the man page on printf:
eE The argument is printed in the style e `[-d.ddd+-dd]' where is one digit before the decimal point and the number after is equal to the precision specification for the argument; when the precision is missing, 6 digits are produced.
So you would want a format something like: %.4e
float n = 6.4516e-10;
n = n * 10;
NSLog(#"n: %.4e", n);
2011-08-29 07:36:38.158 Test[39477:707] n: 6.4516e-09
i have a one application i know The range of a double is **1.7E +/- 308 (15 digits).**but in my application i have to devide text box 's value to 100.0 my code is
double value=[strPrice doubleValue]/100.0;
NSString *stramoount=[#"" stringByAppendingFormat:#"%0.2f",value ];
when i devide 34901234566781212 by 100 it give me 349012345667812.12 but when i type
349012345667812124 and devide by 100 it give me by 100 it give me 3490123456678121.00 which is wrong whether i change datatype or how can i change my code
The number 349012345667812124 has 18 decimal digits. the double format only provides slightly less than 16 decimal digits of precision (the actual number is not an integer because the format's binary digits do not correspont directly to decimal ones). Thus it is completely expected that the last 2 or 3 digits cannot be represented accurately, and it already happens when the literal "349012345667812124" is parsed to the double format, before any calculations happen.
The fact that you get the expected result with the number 34901234566781212 means nothing; it just happens to be close enough to the nearest value the double format can represent.
To avoid this problem, use the NSDecimal or NSDecimalNumber types.
Use
NSDecimalNumber * dec=[[NSDecimalNumber decimalNumberWithString:value.text locale: [NSLocale currentLocale]] decimalNumberByDividingBy:[NSDecimalNumber decimalNumberWithString:#"100" locale:[NSLocale currentLocale]]];
NSLog(#"%#",dec);
instead of Double
I want to convert the decimal number 27 into binary such a way that , first the digit 2 is converted and its binary value is placed in an array and then the digit 7 is converted and its binary number is placed in that array. what should I do?
thanks in advance
That's called binary-coded decimal. It's easiest to work right-to-left. Take the value modulo 10 (% operator in C/C++/ObjC) and put it in the array. Then integer-divide the value by 10 (/ operator in C/C++/ObjC). Continue until your value is zero. Then reverse the array if you need most-significant digit first.
If I understand your question correctly, you want to go from 27 to an array that looks like {0010, 0111}.
If you understand how base systems work (specifically the decimal system), this should be simple.
First, you find the remainder of your number when divided by 10. Your number 27 in this case would result with 7.
Then you integer divide your number by 10 and store it back in that variable. Your number 27 would result in 2.
How many times do you do this?
You do this until you have no more digits.
How many digits can you have?
Well, if you think about the number 100, it has 3 digits because the number needs to remember that one 10^2 exists in the number. On the other hand, 99 does not.
The answer to the previous question is 1 + floor of Log base 10 of the input number.
Log of 100 is 2, plus 1 is 3, which equals number of digits.
Log of 99 is a little less than 2, but flooring it is 1, plus 1 is 2.
In java it is like this:
int input = 27;
int number = 0;
int numDigits = Math.floor(Log(10, input)) + 1;
int[] digitArray = new int [numDigits];
for (int i = 0; i < numDigits; i++) {
number = input % 10;
digitArray[numDigits - i - 1] = number;
input = input / 10;
}
return digitArray;
Java doesn't have a Log function that is portable for any base (it has it for base e), but it is trivial to make a function for it.
double Log( double base, double value ) {
return Math.log(value)/Math.log(base);
}
Good luck.
I have a method that receives a number in a NSString format.
I wish to convert this string to a double which I can use to calculate a temperature.
Here's my method.
NSString *stringTemp = text; // text is a NSString
NSLog(#"%#",stringTemp); // used for debugging
double tempDouble = [stringTemp doubleValue];
NSLog(#"%f",tempDouble); // used for debugging
Please note I put the NSLog commands here just to see if the number was correct. The latter NSLog returns a value of 82.000000 etc. (constantly changes as it's a temperature).
Next I wanted to use this double and convert it to a Celsius value. To do so, I did this:
double celsiusTemp = (5 / 9) * (tempDouble - 32);
Doing this: NSLog(#"%d", celsiusTemp); , or this: NSLog(#"%f", celsiusTemp); both give me a value of 0 in the console. Is there any reason why this would be happening? Have I made a stupid mistake somewhere?
Thank you for your help!
Try doing (5.0 / 9.0). If you only use an int to do math where you are expecting a double to be returned (like 0.55) everything after the decimal place will be lost because the cpu expects an int to be returned.
5 / 9 is the division of two integers, and as such uses integer division, which performs the division normally and then truncates the result. So the result of 5 / 9 is always the integer 0.
Try:
double celsiusTemp = (5.0 / 9) * (tempDouble - 32);
If you evaulate (5/9) as an integer, then it is just 0.