If I define a lazy property with an initializer as in this code. Does initializer have access to X ...

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[alwyn]

If I define a lazy property with an initializer as in this code. Does initializer have access to X and if so how is it accessed?

-.kt

[Vladyslav Sitalo]

if initializer is a lambda it can capture the local closure, so it’d have an implicit this which would refer to an instance of X

[alwyn]

Would something like:
val initializer = fun () { return something; }
be interpreted as a lambda function?

[alwyn]

I suspect not

[Andreas Sinz]

this is an anonymous function, a lambda looks like { ... }

[Andreas Sinz]

https://kotlinlang.org/docs/reference/lambdas.html#lambda-expressions-and-anonymous-functions

[alwyn]

maybe
val initialzer : () -> X = { }

[Andreas Sinz]

this sounds like an XY-problem, what are you trying to achieve?

[Vladyslav Sitalo]

anonymous function should work fine too, as long as it’s defined in the context of the class

[alwyn]

my reason for not wanting to specify the lambda in the class is that it's quite a bit of code, but then I guess the lambda could just be something like { t -> initializer(t) }

[alwyn]

although I'm not sure t in this case refers to the enclosing type

[Vladyslav Sitalo]

{ initializer(this) } should work fine for you.

[alwyn]

I have just enough knowledge to be dangerous 🙂

[alwyn]

I will try that, thanks!

[Vladyslav Sitalo]

also you can define initializer as extension function if you want to have an implicit this

[Vladyslav Sitalo]

though in both cases you won’t have access to non-public stuff

[alwyn]

makes sense

[alwyn]

Thanks again