This is a question on Type Inference that I posted on StackOverflow. First guess the answer and then...

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[Vishnu Haridas]

This is a question on Type Inference that I posted on StackOverflow. First guess the answer and then look at the SO answer.

I understand that Kotlin is a statically-typed language, and all the types are defined/inferred at the compile time itself.

Here is a when expression that returns different types:

fun main(){

        val x = readLine()?.toInt() ?: 0

        val y = when(x){
            1 -> 42
            2 -> "Hello"
            else -> 3.14F
        }

        println(y::class.java)
}

During runtime (Kotlin 1.3.41 on JVM 1.8) this is the output:

When x = 1, it prints class java.lang.Integer
When x = 2, it prints class java.lang.String
Otherwise, it prints class java.lang.Float

When does the compiler determine the type of y? Or, how does the compiler infers the type of y during compile-time?

Question on StackOverflow: https://stackoverflow.com/q/58479824/816416

[Pablichjenkov]

I think the compiler determines y class to be the upper bound class that cover all the when branches return types.
Or in other words, the super class that is common to all branches result.
Now KClass<T>.java is a runtime function call. So it is acting on the actual returned type instance.

[Animesh Sahu]

Using reflection, you get actual class of which instance is of, while in runtime type maybe any supertype, you can declair any integer while storing it as Any type, you cannot do any operation with int since runtime doesn't know about it but if you try using reflection you can actually get the real name of the class of which the instance is of.

[Maxim Kizub]

Compile-time type of y is Any. But runtime type can be anything that extends Any. You print runtine type, not compile-time type.

[Stephan Schroeder]

so just to be very explicit: when doesn’t return different types, it returns a value of typ Any. So you wouldn’t be able to call y.substring(..) without casting y even if x=2.

[Vishnu Haridas]

@Stephan Schroeder AFAIK, when returns the nearest supertype of all branches. In this case, Any.

If we consider classes class Foo, class Bar: Foo, class Baz: Foo, and a when returns Bar and Baz in different branches, then the inferred type will be Foo.

[Vishnu Haridas]

@Stephan Schroeder Also you are right, we must cast y to String before we call y.subString()