Scala中的字符串之间的差异,同时将二进制字符串转换为整数

时间:2016-08-31 21:08:25

标签: java string scala

我有一个问题,我想将0和1的列表转换为整数。

scala> val bar=List('0','1','1').toString
bar: String = List(0, 1, 1)

scala> val foo="011"
foo: String = 011

scala> Integer.parseInt(foo,2)
res1: Int = 3

scala> Integer.parseInt(bar,2)
java.lang.NumberFormatException: For input string: "List(0, 1, 1)"
        at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
        at java.lang.Integer.parseInt(Integer.java:580)
        at .<init>(<console>:9)
        at .<clinit>(<console>)
        at .<init>(<console>:7)
        at .<clinit>(<console>)
        at $print(<console>)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
        at java.lang.reflect.Method.invoke(Method.java:498)
        at scala.tools.nsc.interpreter.IMain$ReadEvalPrint.call(IMain.scala:734)
        at scala.tools.nsc.interpreter.IMain$Request.loadAndRun(IMain.scala:983)
        at scala.tools.nsc.interpreter.IMain.loadAndRunReq$1(IMain.scala:573)
        at scala.tools.nsc.interpreter.IMain.interpret(IMain.scala:604)
        at scala.tools.nsc.interpreter.IMain.interpret(IMain.scala:568)
        at scala.tools.nsc.interpreter.ILoop.reallyInterpret$1(ILoop.scala:760)
        at scala.tools.nsc.interpreter.ILoop.interpretStartingWith(ILoop.scala:805)
        at scala.tools.nsc.interpreter.ILoop.command(ILoop.scala:717)
        at scala.tools.nsc.interpreter.ILoop.processLine$1(ILoop.scala:581)
        at scala.tools.nsc.interpreter.ILoop.innerLoop$1(ILoop.scala:588)
        at scala.tools.nsc.interpreter.ILoop.loop(ILoop.scala:591)
        at scala.tools.nsc.interpreter.ILoop$$anonfun$process$1.apply$mcZ$sp(ILoop.scala:882)
        at scala.tools.nsc.interpreter.ILoop$$anonfun$process$1.apply(ILoop.scala:837)
        at scala.tools.nsc.interpreter.ILoop$$anonfun$process$1.apply(ILoop.scala:837)
        at scala.tools.nsc.util.ScalaClassLoader$.savingContextLoader(ScalaClassLoader.scala:135)
        at scala.tools.nsc.interpreter.ILoop.process(ILoop.scala:837)
        at scala.tools.nsc.interpreter.ILoop.main(ILoop.scala:904)
        at xsbt.ConsoleInterface.run(ConsoleInterface.scala:62)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
        at java.lang.reflect.Method.invoke(Method.java:498)
        at sbt.compiler.AnalyzingCompiler.call(AnalyzingCompiler.scala:101)
        at sbt.compiler.AnalyzingCompiler.console(AnalyzingCompiler.scala:76)
        at sbt.Console.sbt$Console$$console0$1(Console.scala:22)
        at sbt.Console$$anonfun$apply$2$$anonfun$apply$1.apply$mcV$sp(Console.scala:23)
        at sbt.Console$$anonfun$apply$2$$anonfun$apply$1.apply(Console.scala:23)
        at sbt.Console$$anonfun$apply$2$$anonfun$apply$1.apply(Console.scala:23)
        at sbt.Logger$$anon$4.apply(Logger.scala:85)
        at sbt.TrapExit$App.run(TrapExit.scala:248)
        at java.lang.Thread.run(Thread.java:745)

解析函数适用于“011”,但不适用于将等效List转换为String的情况。如何将二进制列表转换为整数?

2 个答案:

答案 0 :(得分:3)

toString上调用List会为您提供一个由List(...)构成的字符串,其中包含所述toString的逗号分隔的List ed元素。请注意bar的原始定义如何返回:

bar: List(0, 1, 1)

您正在寻找mkString(在这种情况下没有参数),它会将toString的所有List ed元素连接成一个字符串。

scala> val bar = List('0','1','1').mkString
bar: String = 011

scala> Integer.parseInt(bar, 2)
res0: Int = 3

答案 1 :(得分:2)

调用toString实际上会为您提供描述 List的字符串,而不是仅由List元素组成的实际字符串。您想要mkString(在工作表中):

val a: String = List("0","1","1").toString
a  // res0: String = List(0, 1, 1)
val b: String = List("0","1","1").mkString
b  // res1: String = 011
Integer.parseInt(b, 2)  // res2: Int = 3

任何时候你看到toString假设它提供了一个人类可读的对象版本,它正好描述了该对象是什么。在这种情况下,您需要一个机器可读的Integer.parseInt()版本来处理。