Sorry if this is a noob question, but why do reifi...

# compiler

[Robert Menke]

Sorry if this is a noob question, but why do reified generics require the function definition to be inline?

[simon.vergauwen]

Because inlining is exactly what the compiler does. There is not such thing as reified generics on the JVM so the compiler inlines the generic param.

i.e. inline fun <reified A> fromJson(json: String) = Gson().fromJson(json, A::class.java) when called as fromJson<Person>(json) the compiler simply turns it into Gson().fromJson(json, Person::class.java).

So you could call it reifying generics by inlining.

[jw]

reified just means that the type parameter is resolvable at compile-time. you don't need inline for this, the compiler could rewrite the signature to accept a KClass, for example, and synthesize it at the callsite.

[jw]

inline is just the current implementation where the function body is copy/pasted and the type parameters which are guaranteed to be resolvable to a class at compile-time are replaced with references to said resoled class

[Robert Menke]

inline is just the current implementation

Gotcha I kind of assumed as much I just didn’t know if there was some underlying reason I was missing

[simon.vergauwen]

@jw are there any other options available on the JVM?

[Derek Berner]

the compiler could rewrite the signature to accept a KClass, for example, and synthesize it at the callsite.

Why isn't this the way the Kotlin compiler does it?

[Derek Berner]

As it's currently implemented, I can never use a KClass reference to call a reified function - it has to be constant

[Derek Berner]

Such an implementation as you describe would also allow for reified types in class definitions.