java - 如何在 Java 中将 double 轮次显式转换为 int 轮次？

Question

System.out.println((int)(99.9999999999999));

返回 99

System.out.println((int)(99.999999999999999999999999999999999999999));

返回 100

你能解释一下为什么吗？

Answer 1

double 文字 99.9999999999999 可以表示为小于 100 的 double，因此强制转换to int 截断小数部分，99 是结果。

double 字面值 99.99999999999999999999999999999999999999 具有最接近 100 的实际值，转换为 int 很简单 100 这里。

System.out.println(99.9999999999999);
System.out.println((int)(99.9999999999999));
System.out.println(99.9999999999999 == 100.0);

System.out.println(99.999999999999999999999999999999999999999);
System.out.println((int)(99.999999999999999999999999999999999999999));
System.out.println(99.999999999999999999999999999999999999999 == 100.0);

我在这里使用 100.0 作为 double 文字，比较 2 个 double 值以查看它们是否完全相同 double 值。

打印出来:

99.9999999999999
99
false
100.0
100
true

JLS, Section 3.10.2涵盖浮点文字:

The details of proper input conversion from a Unicode string representation of a floating-point number to the internal IEEE 754 binary floating-point representation are described for the methods valueOf of class Float and class Double of the package java.lang.

和Double.valueOf javadocs状态:

Otherwise, s is regarded as representing an exact decimal value in the usual "computerized scientific notation" or as an exact hexadecimal value; this exact numerical value is then conceptually converted to an "infinitely precise" binary value that is then rounded to type double by the usual round-to-nearest rule of IEEE 754 floating-point arithmetic, which includes preserving the sign of a zero value.

(强调我的)

也就是说，该值被四舍五入以产生实际的 double 值。