This question already has answers here:
Double calculation producing odd result [duplicate]
(3 answers)
Closed 7 years ago.
I have a piece of code as below.
String wtf = "8.40";
float ft = Float.parseFloat(wtf); //8.4
ft *= 100.0F;
the value of "ft" above is coming as 839.99994
I expected output as 840.00000
How can I correct my code so that it gives me 840.00000 as output
String wtf = "8.40";
double ft = Double.parseDouble(wtf); //8.4
ft *= 100.0F;
There's is a lot of rounding when you use float. This is why you get this "unexpected" value. And I use parseDouble because the cost is less.
I suggest using BigDecimal for arithmetic and calculations.
BigDecimal BDa = new BigDecimal("8.40");
BigDecimal BDc = BDa.multiply(new BigDecimal("100.0"));
Use Double.valueOf(wtf) instead of Float.parseFloat(wtf)
Related
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 1 year ago.
When I divided the large number into a small number then the division is correct but when I write the small number to divide a large number the answer returns wrong. In my scenario, the small number always be first. here is my code this code return 7.4074074074074075E-6 but the correct result is 0.0000074074.
double itf = 0.0;
double a = 4.0;
double b = 540000;
itf = a / b;
Log.i(TAG, "savedata: outputvalue=" + itf);
BigDecimal a = new BigDecimal("4");
BigDecimal b = new BigDecimal("540000");
// 0.0000074074
a.divide(b, MathContext.DECIMAL128);
You should use a decimal type. double is outside the scope of support
This question already has answers here:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 2 years ago.
double double1 = 0.174;
double double2 = 0.175;
double diff = Math.abs(double1 - double2);
diff returns 0.0010000000000000009
Now I type:
double double1 = 3.174;
double double2 = 3.175;
double diff = Math.abs(double1 - double2);
I am expecting diff to return the same result, but it returns 9.999999999998899E-4.
Is there a reason for this behaviour?
can use BigDecimal
BigDecimal decimal1=BigDecimal.valueOf(1.744);
BigDecimal decimal2=BigDecimal.valueOf(1.745);
BigDecimal result=decimal2.subtract(decimal1);
double diff=Math.abs(result.doubleValue());
This question already has answers here:
Retain precision with double in Java
(24 answers)
Is floating point math broken?
(31 answers)
Closed 4 years ago.
Eclipse gives the wrong result when trying to calculate the sum of two floats.
In my code, there are 2 float variables: float from = 0.025 and float to = 1.
Then result has double variable: double value = 7 * from / to.
Eclipse compiler shows: value = 0.174999997019767760
In excel calculator, this result was value = 0.175
How can I solve this an issue?
This is simply due to your Java program not rounding the result the same way the excel calculator does. This is a result of the way computers handle floating point arithmetic. You have two options: round the result, or use the java BigDecimal class. If you want to round the result, you can use:
float from = 0.025f;
float to = 1;
double value = 7 * from / to;
DecimalFormat ds = new DecimalFormat("#.###");
double rounded = Double.parseDouble(ds.format(value));
System.out.println(rounded);
If you would rather not have to round, you can use the BigDecimal class.
Java BigDecimal
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 4 years ago.
I want to calculate the of a number I receive in BigDecimal format :
BigDecimal number1 = new BigDecimal("17");
int percentage = 50;
BigDecimal percentageAmount = number1.multiply(new BigDecimal(percentage/100));
but I got a 0 !
Cast the divided result to double. The integer division is returning zero as expected. This should work.
BigDecimal percentageAmount = number1.multiply(new BigDecimal((double)percentage/100));
Or, make the 100 to 100.0.
BigDecimal percentageAmount = number1.multiply(new BigDecimal(percentage/100.0));
These solutions would work if the number is small as you have used. But these solutions won't give the precise results when the number is big. This would be the best approach for avoiding the precision error:
BigDecimal percentageAmount = number1.multiply(BigDecimal.valueOf((double)percentage/100));
This question already has answers here:
Round a double to 2 decimal places [duplicate]
(13 answers)
Closed 9 years ago.
I have a Double value Double val = 49.569632
How can I roundup the val to get 49.57
You can use the DecimalFormat.
double d = 4.569632;
DecimalFormat df = new DecimalFormat("#.##");
System.out.print(df.format(d));
Or you can use the below method as mentioned in this answer as Luiggi Mendoza suggested.
public static double round(double value, int places) {
if (places < 0) throw new IllegalArgumentException();
BigDecimal bd = new BigDecimal(value);
bd = bd.setScale(places, BigDecimal.ROUND_HALF_UP);
return bd.doubleValue();
}
A simple way to round is when printing
double val = 49.569632; // this should be a primitive, not an object
System.out.printf("%.2f%n", val);
or you can round the value first
double rounded = Math.round(val * 1e2) / 1e2;
System.out.println(rounded);
IMHO Using BigDecimal is slower, more complicated to write and no less error prone than using double if you know what you are doing. I know many developer prefer to use a library than write code themselves. ;)