java:三元运算符(?:)中的奇怪的NullPointerException [英] java: weird NullPointerException in ternary operator (? : )

问题描述

请考虑以下代码片段:

private static void doSomething(Double avg, Double min, Double sd) {
    final Double testMin;
    if (avg != null) {
        testMin = Math.max(min, avg - 3 * sd);
    } else {
        testMin = min;
    }
    System.out.println("testMin=" + testMin);

    final Double verwachtMin = avg != null ? Math.max(min, avg - 3 * sd) : min;
    System.out.println("verwachtMin=" + verwachtMin);
}

据我所知(以及我的 IDE 可以告诉我什么)，变量 testMin 和 verwachtMin 应该是等效的.

As far as I know (and for what my IDE can tell me), the variables testMin and verwachtMin should be equivalent.

如您所料，我宁愿写最后 2 行而不是前 7 行.但是，当我向此方法传递 3 个空值时，我在计算 verwachtMin 变量.

As you might expect, I'd rather write the last 2 lines than the first 7. However, when I pass 3 null values to this method, I get an NPE on the calculation of the verwachtMin variable.

有谁知道这是怎么发生的?三元运算符是否计算第二部分，即使条件不是 true?

Does anyone know how this can happen? Does the ternary operator evaluate the 2nd part, even when the condition is not true?

(Java 版本 1.6.0_21)

(Java version 1.6.0_21)

推荐答案

尝试:

final Double verwachtMin = avg != null ? new Double(Math.max(min, avg - 3 * sd)) : min;

或

final Double verwachtMin = avg != null ? Double.valueOf(Math.max(min, avg - 3 * sd)) : min;

三元运算符的交替边的类型是double和Double，这意味着Double被拆箱为double，然后在赋值时我们有一个从 double 到 Double 的装箱.如果 min 的值为 null 那么拆箱 NPEs.

The types of the alternate sides of the ternary operator were double and Double, which means that the Double gets unboxed to double, and then on assignment we have a boxing from double to Double. If the value of min is null then the unboxing NPEs.

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