英文:
Elvis operator and type casting precedence in Groovy
问题
以下是翻译好的内容：
让我们来看下面的简单表达式：
((Double) null ?: 0).getClass()
结果：
- Groovy 3：
class java.lang.Double - Groovy 4：
class java.lang.Integer
有人知道不同行为的原因吗？
我会说Groovy 4是正确的，因为类型转换在Elvis运算符之前应用。
检查了，但在Groovy 4的发行说明中找不到相关信息：https://groovy-lang.org/releasenotes/groovy-4.0.html
英文:
Let's take the following simple expression:
((Double) null ?: 0).getClass()
Results:
- Groovy 3:
class java.lang.Double - Groovy 4:
class java.lang.Integer
Does anyone know the reason for the different behaviour?
I'd say Groovy 4 is correct since the casting is applied before the Elvis operator.
Checked, but couldn't find anything related in the Groovy 4 release notes: https://groovy-lang.org/releasenotes/groovy-4.0.html
答案1
得分: 0
从文档中可以看出，?: 运算符的优先级比类型转换要低得多。类型转换 (type) 的优先级为1，而 Elvis 运算符 ?: 的优先级为14，所以看起来 Groovy 4 做得对。
https://groovy-lang.org/operators.html#_operator_precedence
在 Groovy 3 的文档中也有相同的说明：
http://docs.groovy-lang.org/docs/groovy-3.0.18/html/documentation/#_operator_precedence
唯一的解释可能是在 Groovy 3 中存在一个未被注意到的简单错误，或者可能是在后续版本的 Groovy 3 中修复的错误，具体取决于你用于测试的版本。即使问题已经修复，也许还值得报告它，这样他们可以编写一个单元测试以便将来能够捕捉到类似的问题，因为它可能曾经被忽略了。
英文:
From the docs the precedence of the ?: is much lower than typecast. Typecast (type) is level 1 precedence and the elvis operator ?: is 14 so it looks like Groovy 4 is doing the right thing.
https://groovy-lang.org/operators.html#_operator_precedence
and in Groovy 3 docs it is documented the same:
http://docs.groovy-lang.org/docs/groovy-3.0.18/html/documentation/#_operator_precedence
The only way I might explain it is simple bug in Groovy 3 that went unnoticed, or maybe a bug that was fixed in later versions of Groovy 3 depending on the version you're using to test it. It might be worth it to report it even though its fixed just so they could write a unit test to catch it in the future since it possibly went unnoticed.
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