Java精度损失大一倍

Question

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

打印1.0715086071862673E301

我尝试过的事情

我试图通过使用BigDecimal类扩展此数字：

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

这只是打印：

10715086071862673000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

我也尝试使用DecimalFormat：

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

打印相同的东西：

10715086071862673000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

根据Wolfram Alpha的实际答案是

如何打印所有实际值？

Answer 1

您不能像这样混合和匹配Math，原始类型和BigDecimal，如果要获得真正的精度，请仅使用BigDecimal：

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

Answer 2

如果要查看double的实际值，请使用BigDecimal（double）构造函数。 Double.toString(lnumber)舍入。

System.out.println(new BigDecimal((Math.pow(2, 1000)))) 2，1000 System.out.println(new BigDecimal((Math.pow(2, 1000))))输出107150860718626732094842504906000181056140481170553360744375038837035105112493612219831983156958581275946729175531468251872852856923140435984577574698574834563456777482423547542654760768654768768768654768768654654768768654654768768654657768654654654768768654654768768654654654768654678654654654678758659654654654死状态死失

Math.pow（2,1000）可以用double精确表示，因为它是可表示范围内的2的幂。