This question already has an answer here:
What's the difference between double.toStringAsFixed and toStringAsPrecision in dartlang?
(1 answer)
Closed 3 years ago.
How I can display or keep double values to 2 decimal points?
textContent: 'Total Payable: ' + '€'+finalPrice.toString(),
I need to know how to ensure that
double finalPrice;
that finalPrice is always displayed to 2 decimal points. Whether it is converted to a string or not.
I've tried everything and called my MP.
double finalPrice = 0.00;
I noticed that initialising it as 0.00 seems to do something, but I need something a bit more solid.
The expected result is the client should pay for what they've purchased. The actual result is we are charging the customer a bit more than what they've purchased
double d = 1.5124;
String number = d.toStringAsFixed(2); // 1.51
Even if you have
double d = 1.51 // number: 1.51
double d = 1.5 // number: 1.50
double d = 1 // number = 1.00
You can see you will always have 2 decimal places with toStringAsFixed(2).
Related
I am trying to convert a textfield input of military time into a double. Can anyone help me with this? goal would be if someone enters 13:45 then the output would be 13.75.
Divide your input into 2 halves around the :. Then parse each half, which results in separate hours and minutes ints. Add them together, dividing the minutes by 60 to get your intended double output.
String input = "13:45";
String firstHalf = input.substring(0, input.indexOf(':'));
String secHalf = input.substring(input.indexOf(':') + 1);
int hour = int.parse(firstHalf);
int min = int.parse(secHalf);
double output = hour + min/60;
print(output);//13.75
This question already has an answer here:
How to store 1.66 in NSDecimalNumber
(1 answer)
Closed 5 years ago.
I've read a lot that NSDecimalNumber is the best format to use when using currency.
However, I'm still getting floating point issues.
For example.
let a: NSDecimalNumber = 0.07 //0.07000000000000003
let b: NSDecimalNumber = 7.dividing(by: 100) //0.06999999999999999
I know I could use Decimal and b would be what I'm expecting:
let b: Decimal = 7 / 100 //0.07
I'm using Core Data in my app. So I'm stuck with NSDecimalNumber. Unless I want convert a lot of NSDecimalNumbers to Decimals.
Can someone help me get 0.07?
The problem is that you’re effectively doing floating point math (with the problems it has faithfully capturing fractional decimal values in a Double) and creating a Decimal (or NSDecimalNumber) from the Double value that already has introduced this discrepancy. Instead, you want to create your Decimal values before doing your division (or before having a fractional Double value, even if a literal).
So, the following is equivalent to your example, whereby it is building a Double representation (with the limitations that entails) of 0.07, and you end up with a value that is not exactly 0.07:
let value = Decimal(7.0 / 100.0) // or NSDecimalNumber(value: 7.0 / 100.0)
Whereas this does not suffer this problem because we are dividing a decimal 7 by a decimal 100:
let value = Decimal(7) / Decimal(100) // or NSDecimalNumber(value: 7).dividing(by: 100)
Or, other ways to create 0.07 value but avoiding Double in the process include using strings:
let value = Decimal(string: "0.07") // or NSDecimalNumber(string: "0.07")
Or specifying the mantissa/significant and exponent:
let value = Decimal(sign: .plus, exponent: -2, significand: 7) // or NSDecimalNumber(mantissa: 7, exponent: -2, isNegative: false)
Bottom line, avoid Double representations entirely when using Decimal (or NSDecimalNumber), and you won't suffer the problem you described.
I'd like to round my values to the closest of 5 cent for example:
5.31 -> 5.30
5.35 -> 5.35
5.33 -> 5.35
5.38 -> 5.40
Currently I'm doing it by getting the decimal values using:
let numbers = 5.33
let decimal = (numbers - rint(numbers)) * 100
let rounded = rint(numbers) + (5 * round(decimal / 5)) / 100
// This results in 5.35
I was wondering if there's a better method with fewer steps because sometimes numbers - rint(numbers) is giving me a weird result like:
let numbers = 12.12
let decimal = (numbers - rint(numbers)) * 100
// This results in 11.9999999999999
Turns out..it's really simple
let x: Float = 1.03 //or whatever value, you can also use the Double type
let y = round(x * 20) / 20
It's really better to stay away from floating-point for this kind of thing, but you can probably improve the accuracy a little with this:
import Foundation
func roundToFive(n: Double) -> Double {
let f = floor(n)
return f + round((n-f) * 20) / 20
}
roundToFive(12.12) // 12.1
I will use round function and NSNumberFormatter also but slightly different algorithm
I was thinking about using % but I changed it to /
let formatter = NSNumberFormatter()
formatter.minimumFractionDigits = 2
formatter.maximumFractionDigits = 2
//5.30
formatter.stringFromNumber(round(5.31/0.05)*0.05)
//5.35
formatter.stringFromNumber(round(5.35/0.05)*0.05)
//5.35
formatter.stringFromNumber(round(5.33/0.05)*0.05)
//5.40
formatter.stringFromNumber(round(5.38/0.05)*0.05)
//12.15
formatter.stringFromNumber(round(12.13/0.05)*0.05)
Depending on how you are storing your currency data, I would recommend using a dictionary or an array to look up the original cents value, and return a pre-computed result. There's no reason to do the calculations at all, since you know that 0 <= cents < 100.
If your currency is a string input, just chop off the last couple of digits and do a dictionary lookup.
round_cents = [ ... "12":"10", "13":"15", ... ]
If your currency is a floating point value, well, you have already discovered the joys of trying to do that. You should change it.
If your currency is a data type, or a fixed point integer, just get the cents part out and do an array lookup.
...
round_cents[12] = 10
round_cents[13] = 15
...
In either case, you would then do:
new_cents = round_cents[old_cents]
and be done with it.
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.
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.