How to go from `Function<T,R>?` to `((T) -&g...

# getting-started

[aarjav]

How to go from Function<T,R>? to ((T) -> R)? I used an extension property val Function<T,R>.lambda: ((T) -> R) get() = this::apply and use it like func?.lambda

[voddan]

Function<T,R>? to ((T) -> R)? are the same type, you shouldn't need converting one to the other

[aarjav]

Maybe I'm missing something here with generics then. do i need to specify in/out somewhere?

[aarjav]

the reverse also gives similar type mismatch message

fun <T,R> bar(func: ((T) -> R)? = null) {
    foo<T,R>(func)
}

fun <T,R> foo(func: java.util.function.Function<T, R>? = null) {
    bar<T,R>(func)
}

[karelpeeters]

What about bar<T,R>{ func.apply(it) }?