Java 大 double 损失

Question

 double lnumber = Math.pow(2, 1000);

打印 1.0715086071862673E301

我尝试过的事情

我尝试使用 BigDecimal 类来扩展这个数字:

 String strNumber = new BigDecimal(Double.toString(lnumber)).toPlainString();

这只是打印:

10715086071862673000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

我还尝试使用 DecimalFormat:

    DecimalFormat df = new DecimalFormat("#");
    df.setMaximumFractionDigits(0);
    String strNumber = String.valueOf(df.format(lnumber));

打印同样的东西:

10715086071862673000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

根据 Wolfram Alpha 的实际答案是

如何打印所有实际值？

Answer 1

你不能像那样混合和匹配 Math、原始类型和 BigDecimal，如果你想要真正的精度，只需使用 BigDecimal:

BigDecimal value = new BigDecimal(2);
System.out.println(value.pow(1000));