octavian - 1 month ago 9

Scala Question

I have defined an identity function and a composition function:

`def identity[T](x: T): T = x`

def composition[A, B, C](f: A => B, g: B => C)(x: A) = g(f(x))

I am trying to assert that the identity function can be applied on both sides with the same result:

`assert(composition((x: Int) => x + 1, identity)(3) == composition(identity, (x: Int) => x + 1)(3))`

However, I am getting these errors:

`Error:(7, 40) type mismatch;`

found : Nothing => Nothing

required: Int => Nothing

assert(composition((x: Int) => x + 1, identity)(3) == composition(identity, (x: Int) => x + 1)(3));}

And:

`Error:(7, 68) type mismatch;`

found : Nothing => Nothing

required: A => Nothing

assert(composition((x: Int) => x + 1, identity)(3) == composition(identity, (x: Int) => x + 1)(3));}

Why is this the case?

Answer

This is the case because the compiler is having difficulty inferring the types properly, specifically inferring what `A`

is. You can help him by placing `A`

as the first argument and each function in a separate parameter list:

```
def composition[A, B, C](x: A)(f: A => B)(g: B => C) = g(f(x))
```

And now this works as expected:

```
scala> :pa
// Entering paste mode (ctrl-D to finish)
def identity[T](x: T): T = x
def composition[A, B, C](x: A)(f: A => B)(g: B => C) = g(f(x))
// Exiting paste mode, now interpreting.
identity: [T](x: T)T
composition: [A, B, C](x: A)(f: A => B)(g: B => C)C
scala> println(composition(3)((x: Int) => x + 1)(identity) == composition(3)(identity)((x: Int) => x + 1))
true
```

Alternatively, you can specify the type parameter explicitly to help the compiler infer the right type:

```
println(composition((x: Int) => x + 1, identity[Int], 3) ==
composition(identity[Int], (x: Int) => x + 1, 3))
```