How to get kotlin function's caller

Question

With this example:

open class Parent {
    fun some():Parent {
        return this;
    }
}

class A : Parent(){
    val name:String? = null;
}

But then this code results in an error:

val a = A().some().some()
a.name // ERROR

EDITOR NOTE: based on comments of the author to answers below, the question is NOT about referencing a.name but really is about something like "how do I get the instance of the class or its name that first started the chain of method calls". Read all comments below until the OP edits this for clarity.

my final goal is to return caller's type and can call this caller's instance property, no more as , no more override, any idea?

I cleaned up the question and added the open keyword you mentioned in a comment was already there, so people do not miscontrue what the problem might be — Jayson Minard, May 26 '17 at 17:32
Based on your comments you are not saying what your actual question is. You are making it look like you have a problem referencing a property of class `A` but that is not what you are trying to do at all, you should update the question based on your comments. I added an editors note which you should remove and replace with a clarifying message of your own. — Jayson Minard, May 26 '17 at 17:33

score 1 · Answer 1 · edited May 26 '17 at 17:34

1

Just like java, you can use stackTrace's getMethodName(). Refer to the kotlin doc.

edited May 26 '17 at 17:34

Jayson Minard

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answered May 26 '17 at 03:30

LF00

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Avijit Karmakar · Answer 2 · 2017-05-26T04:43:59.570

0

open class Parent{
    open fun foo(): Parent {
        return this;
    }
}

This is your Parent class. Parent class has a method named foo(). foo() is a method of class A which will return the instance of it's own class. We must have to open the class and method because by default their visibility modifier is final.

class A : Parent() {
    override fun foo(): A { return this }
}

This is a class named A which extends Parent class. foo() is a method of class A which will return the instance of it's own class.

We will call it like this:

var a = A().foo().foo()

edited May 26 '17 at 04:43

answered May 26 '17 at 03:44

Avijit Karmakar

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it's ok but, how to use A().foo().foo() ,foo() is parent class's method. – junk May 26 '17 at 04:10
if i won't override function ,is there any way to realize this? – junk May 26 '17 at 04:46
We have to override because the method named foo() is present in both Parent and Child class. – Avijit Karmakar May 26 '17 at 04:49
but when i don't override , A().foo() is worked ,and return A's entity, but cant continue use any function on this entity ,why? – junk May 26 '17 at 04:56
If you use a different name, then you didn't face any problem. – Avijit Karmakar May 26 '17 at 04:58

Dmitrii Nechepurenko · Accepted Answer · 2017-05-26T05:27:38.863

0

Actially your example is working(if you add open keyword because all classes in Kotlin are final by default:

A.kt

 open class A {
    fun some(): A {
        return this
    }
 }

B.kt

 class B : A() {
     val test = "test"
 }

And usage

val tmpB = (B().some().some() as B)
val test = tmpB.test

Edited: It because function some() return parent class which doesn't have child class property. So you need to cast it to child class.(Update code)

edited May 26 '17 at 05:27

answered May 26 '17 at 04:47

Dmitrii Nechepurenko

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i forgot add open ,sorry, but it won't work ,i tried. – junk May 26 '17 at 04:53
@junk Strange. Because I've just compiled and run this code. What is your kotlin version and plugin? – Dmitrii Nechepurenko May 26 '17 at 04:56
version : 1.1.1-release 2016.1.1 IDEA – junk May 26 '17 at 05:01
any idea about this? – junk May 26 '17 at 05:04
No ideas...Try update kotlin then clean and rebuild project. – Dmitrii Nechepurenko May 26 '17 at 05:06
yeah i know this now ,but new question is here: function some() return :A type, but this can't get B's property, i will update my question – junk May 26 '17 at 05:15
but this does not do really what you are asking @junk and only just happens to work for the contrived case. – Jayson Minard May 26 '17 at 17:38

score 0 · Answer 4 · answered May 26 '17 at 05:28

0

Your class always return Parent instance. This class do not have any field with the name name. To do that you have 2 ways:

The first:

open class Parent{
    fun some():Parent{
        return this
    }
}

class A :Parent(){
    val name:String? = null
}

fun main() {
    val a = (A().some().some() as A)
    a.name = "";
}

The second:

open class Parent{
    open fun some():Parent{
        return this
    }
}

class A :Parent(){
    override fun some():A {
        return this
    }

    val name:String? = null
}

fun main() {
    val a = A().some().some()
    a.name = "";
}

answered May 26 '17 at 05:28

Trần Đức Tâm

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I think the second solution's better in your case @junk. – Trần Đức Tâm May 26 '17 at 05:30
my final goal is to return caller's type and can call this caller's instance property, no more as , no more override, any idea? – junk May 26 '17 at 05:44
No! You can't do that. – Trần Đức Tâm May 26 '17 at 05:50
@junk should update the question to reflect what is being asked, it is misleading as written now – Jayson Minard May 26 '17 at 17:33

score -2 · Answer 5 · answered May 26 '17 at 07:06

-2

i have know how to do this:

@Avijit Karmakar

@Trần Đức Tâm

use inline function

 inline fun <reified T:Any> some(sql: String):T {

        return this as T ;
    }

answered May 26 '17 at 07:06

junk

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How to get kotlin function's caller

5 Answers5