scala 将任何字符串转换为 Int
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/20006025/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Cast Any of String to Int
提问by アレックス
I have a variable of type Anywith runtime type of Stringwhich I want to cast to Int:
我有一个类型Any为运行时类型的变量String，我想将其转换为Int：
val a: Any = "123"
If I try to cast in to Int, I'll get an exception java.lang.ClassCastException:
如果我尝试转换为 Int，我会得到一个异常java.lang.ClassCastException：
val b = a.asInstanceOf[Int]
How do I do that then?
那我该怎么做呢？
回答by 0__
Casting doesn't convertyour type, it simply tells the system that you think you are smart enough to know the proper type of an object. For example:
转换不会转换您的类型，它只是告诉系统您认为您足够聪明，可以知道对象的正确类型。例如：
trait Foo
case class Bar(i: Int) extends Foo
val f: Foo = Bar(33)
val b = f.asInstanceOf[Bar] // phew, it works, it is indeed a Bar
What you are probably looking for is to convert a Stringto an Int:
您可能正在寻找的是将 a 转换String为 an Int：
val a: Any = "123"
val b = a.asInstanceOf[String].toInt
Or, since you can call toStringon any object:
或者，因为您可以调用toString任何对象：
val b = a.toString.toInt
You can still get a runtime exception if the string is not a valid number, e.g.
如果字符串不是有效数字，您仍然可以获得运行时异常，例如
"foo".toInt // boom!
回答by flavian
In general you steer clear of class casts and nested try catchblocks in Scala.
通常，您会避开try catchScala 中的类强制转换和嵌套块。
import scala.util.{ Try, Failure, Success };
val x: Any = 5;
val myInt = Try { x.toString.toInt) } getOrElse { 0 // or whatever}; // this is defaulting
val mySecondInt = Try { x.toString.toInt };
mySecondInt match {
case Success(theactualInt) => // do stuff
case Failure(e) => // log the exception etc.
}
回答by djcastr0
scala> val a:Any = "123"
a: Any = 123
scala> val b = a.toString.toInt
b: Int = 123
