你可以做正常的事情(见无论是Rex的或吉日的答案) ,或者你可以:
scala> Array("bob","sue")
res0: Array[String] = Array(bob, sue)
嘿,不公平! REPL印出来真的很棒。
scala> res0.toString
res1: String = [Ljava.lang.String;@63c58252
没有喜悦,直到:
scala> runtime.ScalaRunTime.stringOf(res0)
res2: String = Array(bob, sue)
scala> runtime.ScalaRunTime.replStringOf(res0, res0.length)
res3: String =
"Array(bob, sue)
"
scala> runtime.ScalaRunTime.replStringOf(res0, 1)
res4: String =
"Array(bob)
"
我不知道是否有在REPL宽度设置。更新:没有。它固定在
val maxStringElements = 1000 // no need to mkString billions of elements
但我不会尝试十亿:
scala> Array.tabulate(100)(identity)
res5: Array[Int] = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99)
scala> import runtime.ScalaRunTime.replStringOf
import runtime.ScalaRunTime.replStringOf
scala> replStringOf(res5, 10)
res6: String =
"Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
"
scala> res5.take(10).mkString(", ")
res7: String = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
等待,让我们做的是:
scala> res5.take(10).mkString("Array(", ", ", ")")
res8: String = Array(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
这可能是显而易见的:
scala> var vs = List("1")
vs: List[String] = List(1)
scala> vs = null
vs: List[String] = null
scala> vs.mkString
java.lang.NullPointerException
所以相反:
scala> import runtime.ScalaRunTime.stringOf
import runtime.ScalaRunTime.stringOf
scala> stringOf(vs)
res16: String = null
而且,阵列不需要深从其stringPrefix受益:
scala> println(res0.deep.toString)
Array(bob, sue)
你更喜欢哪一种方法,你可以用它:
implicit class MkLines(val t: TraversableOnce[_]) extends AnyVal {
def mkLines: String = t.mkString("", EOL, EOL)
def mkLines(header: String, indented: Boolean = false, embraced: Boolean = false): String = {
val space = "\u0020"
val sep = if (indented) EOL + space * 2 else EOL
val (lbrace, rbrace) = if (embraced) (space + "{", EOL + "}") else ("", "")
t.mkString(header + lbrace + sep, sep, rbrace + EOL)
}
}
但阵列将需要特殊转换,因为您没有获得ArrayOps:
implicit class MkArrayLines(val a: Array[_]) extends AnyVal {
def asTO: TraversableOnce[_] = a
def mkLines: String = asTO.mkLines
def mkLines(header: String = "Array", indented: Boolean = false, embraced: Boolean = false): String =
asTO.mkLines(header, indented, embraced)
}
scala> Console println Array("bob","sue","zeke").mkLines(indented = true)
Array
bob
sue
zeke
尝试在Java中类似的代码,你会看到,它做同样的事情。 'T []。toString()'不返回包含数组元素的String。 – Jashaszun
我在java中有点生疏:D ...所以,你说没有打印数组的解决方案,但是自己写函数? ... –
dupe http://stackoverflow.com/questions/3328085/scala-printing-arrays和其他人 –