Isn't the argument type co- not contra-variant?

Question

I understand the terms co-variance and contra-variance. But there is one small thing I am unable to understand. In the course "Functional Programming in Scala" on coursera, Martin Ordersky mentions that:

Functions are contravariant in their argument types and co-variant in their return types

So for example in Java, let Dog extends Animal. And let a function be :

void getSomething(Animal a){

and I have the function call as

Dog d = new Dog();
getSomething(d)

So basically what is happeneing is that Animal a = d. And according to wiki covariance is "Converting wider to narrow". And above we are converting from dog to Animal. SO isnt the argument type covariant rather than contravariant?

Answer 1

This is how functions are defined in Scala:

trait Function1 [-T1, +R]  extends AnyRef

In English, parameter T1 is contravariant and result type R is covariant. What does it mean?

When some piece of code requires a function of Dog => Animal type, you can supply a function of Animal => Animal type, thanks to contravariance of parameter (you can use broader type).

Also you can supply function of Dog => Dog type, thanks to covariance of result type (you can use narrower type).

This actually makes sense: someone wants a function to transform dog to any animal. You can supply a function that transforms any animal (including dogs). Also your function can return only dogs, but dogs are still animals.

Answer 2

Converting Dog to Animal is converting narrow to wider, so it's not covariance.