Playing with types. How can I make the default pr...

# getting-started

[Pihentagy]

Playing with types.

How can I make the default prepare to a lambda returning Unit, making the default type of U to Unit?

https://pl.kotl.in/c1FbX3GNm

Untitled

[Joffrey]

You can't, because Rule<String, Int>(String::class, "strRule") wouldn't be valid if prepare returned Unit

[Joffrey]

What you can do instead is declare a Rule subclass that has only one type parameter (not U), and that provides a default prepare.

[Pihentagy]

By default I need no additional info, that's why should the prepare lambda return Unit

[Pihentagy]

So I cannot solve this without the need to explicitly state: I create a Rule without "prepare"?

[Joffrey]

You can also add a convenience factory function named Rule with a single type parameter and a default prepare (you don't actually need a new type):

fun <T: Any> Rule(
    clazz: KClass<T>,
    name: String,
    prepare: T.() -> Unit = {},
    whenTrue: T.() -> Boolean = { true },
    thenDo: T.() -> Unit = {},
): Rule<T, Unit> = Rule(clazz, name, prepare, whenTrue, thenDo)

(Style-wise, it's generally accepted to have a capitalized name for a factory function)

[Joffrey]

You can even make this function inline with reified T and you won't need the clazz parameter anymore.

[Pihentagy]

Wow, and you still can use type parameter if you need to:

val rule = Rule<String>("string rule", action = { trigger.trigger() })

[Joffrey]

Yes, the type parameter is a "more convenient" way to specify the class on the call site. It probably cannot be inferred from those lambdas in your case.

[Pihentagy]

Hm, but this does not compiles:

val rule = Rule<Int>("int rule", prepare = { this * 2 })

[Pihentagy]

https://pl.kotl.in/k0AnjL0F9

[Joffrey]

If you want to use prepare, it means you need to use the real constructor (or declare another factory function overload with 2 type parameters).

[Joffrey]

Using the real constructor means calling it like this:

val rule = Rule(Int:: class, "int rule", prepare = { this * 2 })

[Joffrey]

Using another overload means adding this:

inline fun <reified T : Any, U : Any> Rule(
    name: String,
    noinline prepare: T.() -> U,
    noinline condition: T.(U) -> Boolean = { true },
    noinline action: T.(U) -> Unit = {},
): Rule<T, U> = Rule<T, U>(T::class, name, prepare, condition, action)

And then creating the rule like this:

val rule = Rule<Int, Int>("int rule", prepare = { this * 2 })

[Pihentagy]

Thanks, amazing but almost too high for me.
I let the IDE insert the nolinline word, but what does it do exactly?

[Joffrey]

Here we're using inline on this function for the sole purpose of getting a reified T (so we can grab its class without the need for the caller to pass it). However, the compiler doesn't know that it's "just for this". It believes you want to inline everything, including the lambda parameters. That's a problem for you, because you need to pass those lambdas around (and store them in your Rule class as properties), so they cannot be inlined in the call site. This is what noinline is for. You basically tell the compiler that you don't need/want to inline the lambda parameters of the function.

[Joffrey]

Concretely, from the perspective of the caller, this will forbid stuff that's only possible when a lambda is inlined. For instance, using non-local returns, calling suspend functions, etc. But in your case it's ok to restrict this, non-local returns wouldn't make sense anyway.

[Pihentagy]

Can the compiler infer the U from the this*2 expression?

so type just:

val rule = Rule<Int>("int rule", prepare = { this * 2 })

[Joffrey]

This syntax is not valid, but I believe in Kotlin 1.9.? some inference was added. You can try using the _ placeholder for the second type argument:

Rule<Int, _>("int rule", prepare = { this * 2 })

This didn't work for me on the playground, but it might be a playground bug

[Pihentagy]

Yes, it works! 👍