java - 什么决定了java除法运算中的Nan和Infinity

Question

下面代码的输出让我很困惑。为什么有时是 NaN 而有时是 Infinity？

public static void main (String[] args) {

    double a = 0.0;
    double b = 1.0;
    int c = 0; 
    System.out.println(a/0.0);
    System.out.println(a/0);
    System.out.println(b/0.0);
    System.out.println(b/0);
    System.out.println(c/0.0);
    System.out.println(c/0);
}

输出是:

NaN
NaN
Infinity
Infinity
NaN
Exception in thread "main" java.lang.ArithmeticException: / by zero

这里的决定因素是什么？

Answer 1

这是因为 IEEE 浮点运算标准 (IEEE 754)，这是电气和电子工程师协会 (IEEE) 于 1985 年制定的浮点计算技术标准.

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目的:

The IEEE floating-point standard,.. specifies that every floating point arithmetic operation, including division by zero, has a well-defined result. The standard supports signed zero, as well as infinity and NaN (not a number). There are two zeroes: +0 (positive zero) and −0 (negative zero) and this removes any ambiguity when dividing.

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规则:

In IEEE 754 arithmetic, a ÷ +0 is positive infinity when a is positive, negative infinity when a is negative, and NaN when a = ±0.

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Source