Scala下限类型示例

Question

I have the following code:

class A[+X] {
  def printY[Y >: X](y: Y) = println(y)
}

class F
class G extends F

val f = new F
val g = new G

val a = new A[F]
a.printY(g)

I expect a compilation error in a.printY(g) as g has type G which is not a supertype of F . But in class A I indicated that printY method only accepts supertype of A 's type paramter which is F in my example.

Why it works correctly?

Answer 1

Because g is also an instance of F . Therefore, Y is inferred as F , which is a supertype of F . Ie:

a.printY[F](g)

However, this wouldn't compile:

a.printY[G](g)

Answer 2

Note: if you want a.printY(g) not to compile you need to rewrite the method:

def printY[Y](y: Y)(implicit ev: X <:< Y) = println(y)

This way the compiler infers the type parameter to be G and then checks whether it's a supertype of F instead of looking for a supertype of F which also happens to be an acceptable type for g .

Answer 3

Just try to explain why it works from another two angles.

First, as you know the upper bound is reflexive , for Y >: X, type X or subtype of Y is acceptable. so when you define val a = new A[F] , printY will be like:

def printY[Y >: F](y: Y) = println(y)

when you invoke a.printY(g), printY's type parameter will be inferred as G, which is also a type of F.

Second, for def printY[Y >: F](y: Y) = println(y) , when you pass an instance I of type Y, compiler will try to find the common parent of I and F and make the resultant type as the type parameter of printY, so you can even pass value of String, Int to printY.

a.printY[Any]("xxx")
a.printY[Any](3)