This question already has answers here:
The accuracy of a double in general programming and Java
(2 answers)
Closed 2 years ago.
I am trying rounding and format techniques in Java as I have to keep my values in database with some specific formats.
My database column has the data precision of 18,4 which means 14 integers and 4 decimal places at max.
Now I am trying max possible value case which is 99999999999999.9999. When I execute the below code, I am getting a rounded value of 1000000000000000. However, I want to store the exact value provided.
Can anyone suggest to keep it as it is in java variable?
DecimalFormat df2 = new DecimalFormat( "#.####" );
df2.setRoundingMode(RoundingMode.CEILING);
double number = Double.parseDouble("99999999999999.9999");
System.out.println("Format: " + df2.format(number));
You can use BigDecimal
BigDecimal number = new BigDecimal("99999999999999.9999");
Related
This question already has answers here:
Format of BigDecimal number
(4 answers)
How do I format a number in Java?
(9 answers)
Is floating point math broken?
(31 answers)
Closed 3 years ago.
I'm using BigDecimal data type, when I set
new BigDecimal(21.30);
then, I returned it as xml source, it shows as
21.30000000000710522735760100185871124467578125
Another number
new BigDecimal(23.11)
returned
23.1099999994315658113818187512921995859375
I want to show results with the same decimals as I set on create.
The results of this constructor can be somewhat unpredictable. One
might assume that writing new BigDecimal(0.1) in Java creates
a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625.
This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1,
appearances notwithstanding.
If you check the BigDecimal java doc you will find out that you will have to use the constructor which takes a String parameter to end up with the behavior you need.
This question already has answers here:
Format double value in scientific notation
(4 answers)
How to round a number to n decimal places in Java
(39 answers)
Closed 6 years ago.
I have been working with big numbers (with more that 20 digits) and for that reason I use Double . The form I get the numbers is like this 8.653762536765E28.
What I want to do is just display the first 2 decimal digits. I want it like 8.65E28.
I tried to find about formatting double values but I wan't able to do it. The result I was getting was 86537...12312.00 .
What do you thing is a good approach for this case? How can I manage only the digits in front of the E (the base) and not the whole number?
I think this is what you want to achieve:
double d = 8.653762536765E28;
DecimalFormat df = new DecimalFormat("0.00E0");
System.out.print(df.format(d));
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 8 years ago.
In Java, for the following code
Double d = 2.0-1.1;
System.out.println(d);
The result is
0.8999999999999999
If the program is dealing with sensitive information such as precentile/money or cents how do I solve this problem?
I also tried the following piece of code:
new BigDecimal(d)
which outputs
0.899999999999999911182158029987476766109466552734375
What should I do to get 0.90 for the above case?
Since double cannot accurately represent the result of 2 - 1.11, the precision has already been lost by the time the constructor is used.Therefore you need to chain BigDecimal using the String based constructor
BigDecimal result = new BigDecimal("2").subtract(new BigDecimal("1.1"));
The Standard Reference is What Every Computer Scientist Should Know About Floating-Point Arithmetic
A more digestable read: Floating Point Numbers
This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 9 years ago.
In this program I'm printing a value from a calculation of type double to the screen.But at present the calculation is giving 14 decimal places.My question is,is there a facility in Java to wrap the output statement in that could specify the amount of decimal places?For example round(mark1,2)
The way it is printed at present is like this:
double markOne = intent.getDoubleExtra("number1", 0);
result1.setText(String.valueOf(markOne)+"mm");
Is it possible to wrap the setText in a Java method or would I have to create a custom format?Could someone give me an example of this with my code? Thanks
You could use a decimal formatter
DecimalFormat df = new DecimalFormat("#.##");
System.out.print(df.format(d));
I think it's better to use BigDecimal type instead of Double.
You could do
BigDecimal myValue= new BigDecimal(12.3577);
myValue= myValue.setScale(2, BigDecimal.ROUND_HALF_UP);
And myValue will be 12.36
I hope this will help
Try this:
int decimalPlaces = 2;
markOne = double(int(markOne*Math.pow(10, decimalPlaces)))/Math.pow(10, decimalPlaces);
This question already has answers here:
How to print formatted BigDecimal values?
(7 answers)
Closed 9 years ago.
I have a BigDecimal whose value I'd like to convert to a string and NOT lose any precision.
The format I'd like to use is ###.###,## (comma for thousands, period for decimals).
The only way I've made this work is using
DecimalFormat formatter = new DecimalFormat("###.###,##");
formatter.format(bd.doubleValue());
... but I'm afraid I might lose information this way, and precision is a must because I'm dealing with currency (every penny counts).
Additional information: I will only be dealing with sums of up to 1 million if that is of any help.
To be on the safe side, you could multiple your BigDecimal by 100.
Then get its intValue -> say N. Then get N/100 and N%100.
This way you cannot lose precision (N <= 100 million cannot overflow int).