Question

为什么无法访问String值？

我希望s1是“a”而不是它的Ljava.lang.String; @ d70d7a？

 val it = Iterator("(a,((a,b),1.0))")             //> it  : Iterator[String] = non-empty iterator

 val s1 = it.next.replace("(" , "").replace(")" , "").split(",").toString.split(",")
                                                  //> s1  : Array[String] = Array([Ljava.lang.String;@d70d7a)
 println("s1 is "+s1(0))                          //> s1 is [Ljava.lang.String;@d70d7a

Answer 1

让我们逐个命令：

val it = Iterator("(a,((a,b),1.0))")
// here we got iterator on one String

val s1 = it.next // "(a,((a,b),1.0))"
        .replace("(" , "") // "a,a,b),1.0))"
        .replace(")" , "") // "a,a,b,1.0"
        .split(",") // multiple lines in array: "a", "a","b","1.0"
        .toString // Array[String].toString returns what you got: Ljava.lang.String;@d70d7a
        .split(",") // one String (because there's no "," signs)

也许您应该在toList之前运行toString，因为toString的定义方式是您希望在List的此实现中定义它：

val s1 = ...
        .split(",")
        .toList
        .toString
        ...

也许您应该查看Java: split() returns [Ljava.lang.String;@186d4c1], why?以获得澄清。

Answer 2

.split(",")使得数组和.toString无法处理数组，之后您再次按.split(",")分割我认为没有用。

您还可以使用replaceAll代替多个replace

scala>  val it = Iterator("(a,((a,b),1.0))")         
it: Iterator[String] = non-empty iterator

scala>  val s1 = it.next.replaceAll("[()]" , "").split(",")
s1: Array[String] = Array(a, a, b, 1.0)

scala> println("s1 is "+s1(0))    
s1 is a

无法访问String值

2 个答案: