val a=10; val b=if(a==5) println("hello") else (1,"hi")

What will be return type inferred in this case? And How?

CodePudding user response：

Presumably you're asking for the type of b, which in this case is Any.

if (a == 5) println("hello") else (1, "hi")

gets typed as follows:

• For an if expression, the type is the least upper bound of the two possible branches: if the consequent expression (println("hello") in this case) has type A and the alternative expression ((1, "hi")) has type B, then it is the type C such that A and B are both non-strict subtypes of C and there is no type D which is a strict subtype of C where A and B are non-strict subtypes of D (non-strict means that we can consider a type to be a subtype of itself; strict means we can't).

• The type of println("hello") is Unit

• The type of (1, "hi") is a Tuple2[Int, String]

• Unit's supertypes are AnyVal and Any

• Tuple2's supertypes are Serializable, Product2[Int, String], Product, Equals, AnyRef, and Any (technically, Product2's and Tuple2's covariance means that there are some more supertypes (e.g. Tuple2[AnyVal, String], Tuple2[Int, AnyRef], Tuple2[Any, Any]), but those don't end up being relevant here)

• The least-upper-bound is therefore Any, so the type of the if expression and thus of b is Any

You can demonstrate this in a REPL, e.g. sbt console

scala> :paste
// Entering paste mode (ctrl-D to finish)

val a=10; val b=if(a==5) println("hello") else (1,"hi")

// Exiting paste mode, now interpreting.

a: Int = 10
b: Any = (1,hi)

Note that even though the predicate a == 5 will never be true and thus the consequent branch will never be taken, the typer does not take that into account.

Note also that if you had a method with that body:

def someMethod = {
  val a = 10
  val b = if(a==5) println("hello") else (1,"hi")
}

The result type of that method would be Unit, because the result type is the type of the last expression in the method and an assignment (being a side-effect) has the type Unit.