If I define a lazy property with an initializer as...
# getting-starteda
alwyn
10/21/2018, 7:40 PMIf 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?
v
Vladyslav Sitalo
10/21/2018, 7:43 PMif
initializerthisXa
alwyn
10/21/2018, 7:46 PMWould something like:
val initializer = fun () { return something; }
be interpreted as a lambda function?
alwyn
10/21/2018, 7:48 PMI suspect not
a
Andreas Sinz
10/21/2018, 7:48 PMthis is an anonymous function, a lambda looks like
{ ... }Andreas Sinz
10/21/2018, 7:49 PMa
alwyn
10/21/2018, 7:50 PMmaybe
val initialzer : () -> X = { }
a
Andreas Sinz
10/21/2018, 7:50 PMthis sounds like an XY-problem, what are you trying to achieve?
v
Vladyslav Sitalo
10/21/2018, 7:51 PManonymous function should work fine too, as long as it’s defined in the context of the class
a
alwyn
10/21/2018, 7:51 PMmy 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
10/21/2018, 7:52 PMalthough I'm not sure t in this case refers to the enclosing type
v
Vladyslav Sitalo
10/21/2018, 7:53 PM{ initializer(this) }a
alwyn
10/21/2018, 7:53 PMI have just enough knowledge to be dangerous 🙂
alwyn
10/21/2018, 7:53 PMI will try that, thanks!
v
Vladyslav Sitalo
10/21/2018, 7:53 PMalso you can define initializer as extension function if you want to have an implicit
thisVladyslav Sitalo
10/21/2018, 7:54 PMthough in both cases you won’t have access to non-public stuff
a
alwyn
10/21/2018, 7:54 PMmakes sense
alwyn
10/21/2018, 8:00 PMThanks again
🙂 1
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