Modular•2y ago

Kotlin's extension function in Mojo?

Not so long ago, I've discovered that Kotlin have a possibility to extend already defined classes with a new function. For example:

fun String.getWordsList(): List<String> {
    return this.split(" ")
}

fun main() {
    val text: String = "Hello world!"
    val words = text.getWordsList()
    println(words)
}

fun String.getWordsList(): List<String> {
    return this.split(" ")
}

fun main() {
    val text: String = "Hello world!"
    val words = text.getWordsList()
    println(words)
}

will return [Hello, world!] list. This feature is really useful, as it makes code really cleaner without overriding whole class. I don't asking Modular team to implement it. I'm just curious what others think about that. Is Mojo needs that or static methods is enough? I mean, we can use text.getWordsList() is Kotlin and get_words_list(text) in Mojo and Python. What is better for you?

7 Replies

sora•2y ago

I think Mojo will have Swift style extension. See this previous answer from Modular.

Ilya I. LubenetsOP•2y ago

Great! But is this really better to extend T and use T.do() instead of do(T)? The only usefull approach I see is ability to extend from import. Like if I do from bobby_lib import tensor_utils and somehow run it - I will get a modified Tensor struct with a bunch of new cool and useful functions that bobby wrote

sora•2y ago

By extending a type T, you can make it conform to a trait a posteriori, which free function doesn't grant you.

Ilya I. LubenetsOP•2y ago

You mean that I can to add __hash__ func to existing struct and also add Hashable trait with this approach?

sora•2y ago

If you prefer, you could always use the type as a namespace, and use T.f(t) Yes, tha'ts exactly the point.

Ilya I. LubenetsOP•2y ago

That makes sense, yep

sora•2y ago

Also, conditional conformance. Sometimes you can't put the Hashable on the generic type itself: Tuple[T] is only hashable if T is hashable.

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Kotlin's extension function in Mojo?

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