```fun <T> identity(t: T): T = t fun <T,...

# getting-started

[y]

fun <T> identity(t: T): T = t

fun <T, E> Iterable<T>.allSameValue(f : (T) -> E = ::identity) : E? { /* ... */ }

I'm getting Required: T, Found E for the default value on the parameter. why? what's wrong with E = T?

[Gleb Minaev]

When T = E, everything is OK. But compiler says, "What if E != T? Then ::identity function does not fit the type (T) -> E."

[Gleb Minaev]

And the code that breaks your concept may look like this:

fun main() {
    listOf(1, 2, 3).allSameValue<Int, String>()
}

[y]

oh... yeah.

[Paul Griffith]

e.g. if you don’t use a callable reference but instead a lambda to call identity you have to make an explicit unsafe cast

[y]

the caller can instantiate this with any E and then ::identity doesn't agree with the requirement of returning E

[Szymon Jeziorski]

You could bound generics so that T must be a subtype of E :
fun <T : E, E> Iterable<T>.allSameValue(f: (T) -> E = ::identity): E?
With such signature identity function would always fit into generics bound, and mapping function could be used with its result matching supertypes of T.

Example:

val l = listOf(123)
val x: Number? = l.allSameValue { it.toDouble() }

[y]

@SJ yep but that makes the function less general. I figured out I can just add another overload of `allSameValue`:
fun <T> Iterable<T>.allSameValue(): T? = allSameValue { it }

[Szymon Jeziorski]

Sure, but using signature with T : E and not not passing any arguments would achieve basically the same, since from the default value of of f being identity function, E would just be inferred as T, since indentity takes and returns same generic type:

val l = listOf(1, 2, 3)
val x = l.allSameValue() // inferred as Int?
val z = l.allSameValue<_, Number>() // inferred as Number? because of explicit generics type declaration