当数字超过1234567891123456时，DecimalFormat解析双重错误

Question

I get an unexpected result when I try to parse string 12345678911234567 to double. I use the code as below

    double convertedNum = 0;
convertedNum = new DecimalFormat().parse(12345678911234567).doubleValue();

the converted result was 1.2345678911234568E16 but I expected it was 1.2345678911234567E16, and I also try to use BigDecimal to convert as below

double convertedNum = 0;
BigDecimal bgNum = null;
bgNum = new BigDecimal(12345678911234567);
convertedNum = bgNum.doubleValue();

will get same result, I can't find what cause this. Would someone can give me some suggests, thanks.

Answer 1

My best suggestion is to put the number in quote marks in the BigDecimal constructor - like

new BigDecimal("12345678911234567");

The constructor of BigDecimal with a String argument can deal with numbers of (just about) any reasonable size.

Not all large numbers can be stored correctly in a double . It seems that the number you're trying to represent may fall in between two double values. But BigDecimal will certainly be able to represent it correctly.

Update

To clarify my last paragraph, there is no double equal to 12345678911234567 . There is a double equal to 12345678911234566 , but the next double is equal to 12345678911234568 . So any attempt whatsoever to store 12345678911234567 in a double will result in rounding up to 12345678911234568 .