scala – 为什么链接匹配表达式不能编译?
发布时间:2020-12-16 18:35:34 所属栏目:安全 来源:网络整理
导读:参见英文答案 Pattern Match “return” value????????????????????????????????????2个 链接匹配表达式无法编译. val x = Array("abc","pqr")x match { case Array("abc",_*) = Some("abc is first") case Array("xyz",_*) = Some("xyz is first") case _ =
参见英文答案 >
Pattern Match “return” value????????????????????????????????????2个
链接匹配表达式无法编译. val x = Array("abc","pqr") x match { case Array("abc",_*) => Some("abc is first") case Array("xyz",_*) => Some("xyz is first") case _ => None } match { case Some(x) => x case _ => "Either empty or incorrect first entry" } 虽然以下编译正常: (x match { case Array("abc",_*) => Some("xyz is first") case _ => None }) match { case Some(x) => x case _ => "Either empty or incorrect first entry" } 为什么后面的版本(第一个匹配表达式在paranthesis中)编译正常而前一个版本没有? 解决方法
如果允许,你不能这样做:
scala> List(1,2,3) last match { case 3 => true } warning: there were 1 feature warning(s); re-run with -feature for details res6: Boolean = true 也就是说,如果它是中缀表示法,那么左边的东西不能是后缀. 禁止中缀匹配允许后缀scrutinee. 该表达式以自然方式解析 (List(1,3) last) match { case 3 => true } 也就是说,如果后缀表示法是自然的而不是邪恶的. 功能警告用于import language.postfixOps.也许关闭该功能,善良的捍卫者将愿意接受import language.infixMatch. 考虑要匹配的语法兄弟的构造,没有parens是不可混合的: scala> if (true) 1 else 2 match { case 1 => false } res4: AnyVal = 1 // not false scala> (if (true) 1 else 2) match { case 1 => false } res1: Boolean = false 要么 scala> throw new IllegalStateException match { case e => "ok" } <console>:11: error: type mismatch; // not "ok",or rather,Nothing found : String("ok") required: Throwable throw new IllegalStateException match { case e => "ok" } ^ scala> (throw new IllegalStateException) match { case e => "ok" } java.lang.IllegalStateException (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |