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- o
orangy
09/10/2016, 9:26 PMAlso, if you have two different overloads like I suggested, you will avoid one
operation infilter
case, which is, I think, a good thing.Any - p
punksta
09/10/2016, 9:29 PMOf course. This is not the point. I can solve it with passingClass<T> - o
orangy
09/10/2016, 9:31 PMThere are many ways to solve the problem 🙂 But some of them do not require any new features in a language and can work as you want them to work right now. - p
punksta
09/10/2016, 9:54 PMOf course. Good luck with working on v1.1 simple smile - z
zeesh49
09/11/2016, 4:01 PMim searching for the '$it' variable/value i keep seeing when I debug code, but I cant find it in kotlin-lang learn page... - z
zeesh49
09/11/2016, 4:01 PMany idea what it is? - z
zeesh49
09/11/2016, 4:02 PMi need to print the type of the '$it' variable/value, how can i do that? - i
ilya.gorbunov
09/11/2016, 4:18 PM@zeesh49 Do you mean
that is default name of a single parameter? You can treat it as usual parameter/variable. More about it in https://kotlinlang.org/docs/reference/lambdas.htmlit - i
ilya.gorbunov
09/11/2016, 4:20 PMTo get the runtime type of an object, you can use
extension property like this:javaClassit.javaClass - z
zeesh49
09/11/2016, 4:21 PMActually there is a foreach loop on an arraylist, and each object in the list has a type which I want to print - z
zeesh49
09/11/2016, 4:21 PM@ilya.gorbunov: - z
zeesh49
09/11/2016, 4:25 PM@ilya.gorbunov The object name is generator, when I inspect it, I see '$ it' whose type I want to print. generator.tostring doesn't work. - i
ilya.gorbunov
09/11/2016, 4:27 PM@zeesh49 did you try inspecting
extension property?javaClass - z
zeesh49
09/11/2016, 4:28 PMlet me try - y
yole
09/12/2016, 9:32 AMcollection.any { it is R && it is F } - k
kirillrakhman
09/12/2016, 9:32 AMthat's not right, is it? - k
kirillrakhman
09/12/2016, 9:33 AMI mean depending on how you interpret the question - y
yole
09/12/2016, 9:33 AMyes - y
yole
09/12/2016, 9:33 AManother interpretation would bec.any { it is R } && c.any { it is F } - k
kirillrakhman
09/12/2016, 9:34 AMthat's how I understood the question - z
zeesh49
09/12/2016, 9:39 AMit says unresolved reference 'it' - z
zeesh49
09/12/2016, 9:41 AMohh got it, sorry, missed the if - m
mzgreen
09/12/2016, 11:25 AMCopy code
When I callfun a() {} fun a(param: Int = 1) {}
the first function is called. Why? And is there any way in this case to use second one with default parameter?a() - m
mzgreen
09/12/2016, 11:25 AMHow does the compiler decide which one to use? In swift there is a compilation error thrown in such case. - d
drfriendless
09/12/2016, 11:26 AMWhy is the first one even there, if you don't want to call it and the second one handles that signature anyway? - d
drfriendless
09/12/2016, 11:27 AM(but yes, a compile error is a good solution) - m
mzgreen
09/12/2016, 11:29 AMI’m not saying it makes any sense. I’ve just found out that it compiles and I’m curious why. - m
mzgreen
09/12/2016, 11:30 AM@yole ? - k
kirillrakhman
09/12/2016, 11:32 AMbecause that's how the overloading works. no parameter is more specific than default parameter