What is the difference between `fun <T> T?.foo(): ...

# getting-started

[Sam Stone]

What is the difference between
fun <T> T?.foo(): T
and
fun <T: Any> T?.foo(): T
? Does one or both guarantee that the output will be non-null?

[Michael Krussel]

In the first example T can be a nullable type and the compile will force you to handle the possible null return from foo.

I think the default type inference will be the non null, but you can force it.

null.foo<Any?>() will have a return type of Any?

[Sam]

In the first example, T can be any type, nullable or otherwise, and additionally, the receiver can be null even if T is given as a non null type. The return type is nullable if and only if T is given as nullable. Whereas in the second example, T must be non null, so the return type must be non null (but the receiver is nullable)

[Sam Stone]

@Sam I don't understand: isn't the type of T the same as the type of the receiver? How can the type be non-nullable if the receiver is nullable? Only by explicitly declaring the type in brackets?

[Michael Krussel]

The receiver will still be nullable with the second version, but T will not be so the return type will not be.

[Sam]

The receiver type is T?, so no, it's not the same as T. The added ? means it can always be null, even if T itself is a non-null type.

[ephemient]

Kotlin will infer the narrowest type reasonable, but with the first, both

val x: String?
x.foo<String>()
x.foo<String?>()

are possible. with the second, only

x.foo<String>()

is possible. (well plus supertypes, so x.foo<Any>() etc. may also be possible)

[Stephan Schröder]

but this version should be equivalent to the second, isn't it?
fun <T> T?.foo(): T & Any
(see definitely-non-nullable-types)