How can I cast a a string to generic number using scala?

Question

I'm trying to convert a generic string to a number using scala

object h extends App {
  def castTo[T](s: String): T = {
    s.asInstanceOf[T]
  }


  print(castTo[Int]("20"))
  print(castTo[Double]("20.1"))
}

the data:

name | value a | "1" b | "2.123" c | "abd"

the usecase: riight now I'm exporting the data to the user a method for each conversion. getNameAsDouble, getNameAsInteger and so forth. I wish to do getNameT to save lot's of code and make it a bit more pretty and easy to read the doc.

so, in case a programmer does : getNameInt i want the program to print in this case: 1 getNameDouble i want the program to print in this case: 2.123

in cpp i could use dynamic_cast. there a way to do so in scala? ( i also tried to do so in java but couldn't find a way)

p.s. i've tried something like this, but i wandered if there is more generic way.

castTo[T] (s:String): T = {
...
case T instance of Integer => s.toInt
case T instance of Long => s.toLong
...
}

What you want to do is to parse the String into a generic Number, not to cast it. BTW, why do you want to do this? String already provides the toInt, toDouble & etc methods, why do you need to encapsulate all of them on a single method? — Luis Miguel Mejía Suárez
– Luis Miguel Mejía Suárez, Commented Dec 10, 2019 at 14:11
i'm exporiting an API to get data from source like this: name | value a | "1" b | "2.123" c | "abd" so the user of the api would do getName[Int](name) — Yair Taboch
– Yair Taboch, Commented Dec 10, 2019 at 14:17
It would be good if you add that to your question with a broader explanation, example of use, expected output and what should happen if the types do not match. — Luis Miguel Mejía Suárez
– Luis Miguel Mejía Suárez, Commented Dec 10, 2019 at 14:39

Luis Miguel Mejía Suárez · Accepted Answer · 2019-12-10 14:46:17Z

3

I believe it would be better if you can expand more on your use case.
But, this should do what you want.

def readAs[T](str: String)(implicit num: Numeric[T]): Option[T] =
  num.parseString(str)

Which you can test like:

readAs[Int]("10")
// res: Option[Int] = Some(10)

readAs[Double]("10")
// res: Option[Double] = Some(10.0)

readAs[Double]("10.0d")
// res: Option[Double] = Some(10.0)

readAs[Int]("10.0d")
// res: Option[Int] = None

readAs[Int]("blah")
// res: Option[Int] = None

answered Dec 10, 2019 at 14:46

Luis Miguel Mejía Suárez

23k4 gold badges31 silver badges50 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Mike Nakis · Accepted Answer · 2019-12-10 14:33:38Z

1

Scala is not javascript. Scala is a real programming language with types. Strong types even. So, it treats conversions between types as what they really are: conversions. Not "casts". So, the string will have to be parsed into a number. And if you wrap this parsing in a function, it is utterly wrong to call this conversion a "cast".

And no, you cannot cast a string to a number in C++ either. Not with a dynamic cast, nor with any other kind of cast. You also have to parse it in C++, because C++ is also a real programming language.

As for simplifying your pattern matching expression, you might be able to first parse the string into a double, and then use a generic cast to convert that double into a number of lesser precision, but I do not have a Scala compiler at hand to prove the concept.

edited Dec 10, 2019 at 14:33

answered Dec 10, 2019 at 14:28

Mike Nakis

63.1k11 gold badges128 silver badges173 bronze badges

1 Comment

Luis Miguel Mejía Suárez Over a year ago

This could be condensed to a comment instead of an answer as it really doesn't answer the question.

Collectives™ on Stack Overflow

How can I cast a a string to generic number using scala?

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related