What is the differences between Int and Integer in Scala?

user_46768 0
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I was working with a variable that I had declared as an Integer and discovered that > is not a member of Integer. Here's a simple example:

scala> i
warning: there were deprecation warnings; re-run with -deprecation for details
res28: Integer = 3

scala> i > 3
<console>:6: error: value > is not a member of Integer
       i > 3
         ^

Compare that to an Int:

scala> j
res30: Int = 3

scala> j > 3
res31: Boolean = false

What are the differences between Integer and Int? I see the deprecation warning but it's unclear to me why it was deprecated and, given that it has been, why it doesn't have a > method.

integer int scala types
5 Answers

Integer gets imported from java.lang.Integer and is only for compatibility with Java. Since it is a Java class, of course it can't have a method called "<". EDIT: You can mitigate this problem by declaring an implicit conversion from Integer to Int.

 implicit def toInt(in:Integer) = in.intValue()

You'll still get deprecation warning though.

Kim_464500

I think the problem you're seeing has has to do boxing/unboxing of value types and the use of the Java class Integer.

I think the answer is here: Boxing and unboxing in Scala. There is no implict unboxing in Scala. You've defined i as the Java class Integer but in the i > 3, the 3 is being treated and an int.

user_1078000

Integer is a Java class, java.lang.Integer. It's different from Java's primitive type int, which is not a class. It can't have < defined, because Java does not allow operators to be defined for classes.

Now, you might wonder why such a type exist at all? Well, primitive types cannot be passed as references, so you can't pass an int to a method expecting java.lang.Object, equivalent to Scala's AnyRef, for example. To do that, you put that int inside an Integer object, and then pass the Integer.

user_530130
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