// ERROR: missing parameter type for expanded function myStrings.foreach(println(_.toString)) It expands to: myStrings.foreach(println(x => x.toString)) You want: myStrings.foreach(x => println(x.toString)) The placeholder syntax for anonymous functions replaces the smallest possible contain...
问Scala代码-错误:缺少参数类型错误EN这段代码有什么问题？根据Picrce的说法：“类型系统是一个可以根据...
scala> l2.fold((x,y) => x-y)<console>:19: error: missing parameter type l2.fold((x,y)=> x-y)^ <console>:19: error: missing parameter type l2.fold((x,y)=> x-y)^ <console>:19: error: missing argument listformethod fold in trait TraversableOnce Unapplied methods are only c...
scala> numbers.foreach((i: Int) => i * 2)什么也没有返回。你可以尝试存储返回值,但它会是 Unit 类型(即void)scala> val doubled = numbers.foreach((i: Int) => i * 2) doubled: Unit = ()filterfilter 移除任何对传入函数计算结果为 false 的元素。返回一个布尔值的函数通常被称为谓词函数[...
问Scala下划线-错误:扩展函数缺少参数类型EN Scala 的类型参数其实意思与 Java 的泛型是一样的，也是...
both value bar in object $iw of type => java.lang.String and value foo in object $iw of type => java.lang.String match expected type String 系统找到两个隐式参数是相同的类型,将无法识别。 子类可以匹配, scala> def sayThings (implicit args : List[Any]) = args.foreach(println(_)) ...
match expected type String 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 系统找到两个隐式参数是相同的类型,将无法识别。 子类可以匹配, AI检测代码解析 scala> def sayThings (implicit args : List[Any]) = args.foreach(println(_)) ...
(Foo,1) scala> val with2: String With Int = With("Bar", 2) with2: With[String,Int] = With(Bar,2) scala> Seq(with1, with2) foreach { w => | w match { | case s With i => println(s"$s with $i") | case _ => println(s"Unknown: $w") | } | } Foo with 1 ...
Type in expressions for evaluation. Or try :help. scala> val it = Iterator(1,2,3) it: Iterator[Int] = non-empty iterator scala> it foreach println 1 2 3 scala> it foreach println scala> it.toArray res2: Array[Int] = Array() ...
def bar(s: String, n: Int): Unit =1to n foreach {i => print(s)} 1. 函数体中出现的{i => print(s)}就是以匿名函数形式定义的闭包。1 to n是1.to(n)的简化形式,然后将闭包作为参数传递给刚创建的Range对象的foreach方法(参数i在闭包的函数体中并没有被使用,仅是为了语法需要)。