问题
I'm trying to implement a new grade rounding to BigDecimal class, and I'm getting a possible bug, must probably I doing something wrong. The code below exposes my problem:
public static void main(String[] args) throws IOException {
BigDecimal valDouble = new BigDecimal(0.35);
valDouble = valDouble.setScale(1, BigDecimal.ROUND_HALF_UP);
System.out.println(valDouble.doubleValue()); // prints 0.3
BigDecimal valString = new BigDecimal(new String("0.35"));
valString = valString.setScale(1, BigDecimal.ROUND_HALF_UP);
System.out.println(valString.doubleValue()); // prints 0.4
}
My doubt here is, is BigDecimal different for double and String constructors?
I can't understand this 'bug', at least, I just used a simple string concat to 'solve' it, as below:
BigDecimal valDouble = new BigDecimal("" + 0.35);
Any idea what could be causing this odd behavior?
回答1:
This isn't a bug. 0.35 as a double literal represents a value that is not exactly equal to 0.35; it's likely something like 0.349999995 or something. So it rounds down.
The String constructor lets you specify 0.35 exactly using "0.35", and that rounds up.
Don't use the double
constructor here; when it matters enough to use BigDecimal
you need to stay out of floating-point land.
回答2:
You don't need to guess what 0.35 is represent as
BigDecimal valDouble = new BigDecimal(0.35);
System.out.println(valDouble);
prints
0.34999999999999997779553950749686919152736663818359375
This will round down to 1 decimal place as 0.3
You don't need to convert to a String, you can use valueOf
BigDecimal valDouble = BigDecimal.valueOf(0.35);
System.out.println(valDouble);
prints
0.35
To round half up to one decimal place you can use
double d = 0.35;
double d1 = Math.round(d * 10) / 10.0;
System.out.println(d1);
prints
0.4
回答3:
0.35 is already inexact, became of the binary radix of FP. Use new BigDecimal("0.35"), which is exact, because of the decimal radix.
来源:https://stackoverflow.com/questions/11368496/java-bigdecimal-bugs-with-string-constructor-to-rounding-with-round-half-up