Is it possible to have an `operator val invoke` property ? The error message seems to suggest its po...

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[spand]

Is it possible to have an operator val invoke property ? The error message seems to suggest its possible but I cant find a way
Modifier 'operator' is not applicable to 'member property without backing field or delegate'

class Foo {
    companion object {
        operator val invoke = {}
    }
}

[Sebastian Sellmair [JB]]

No, How would you access it?

[Sebastian Sellmair [JB]]

What do you want to do with this and why is operator fun invoke not an option? 🤔

[spand]

I feel this is a bit repetetive:

interface Foo {
    var prop1: String
    var prop2: String
    
    companion object {
        operator fun invoke(prop1: String, prop2: String) = object : Foo {
            override var prop1 = prop1
            override var prop2 = prop2
        }
    }
}

So maybe this would be a shorter way to express the same:

interface Foo2 {
    var prop1: String
    var prop2: String

    class Helper(override var prop1: String, override var prop2: String) : Foo2
    
    companion object {
        operator fun val invoke = ::Helper::invoke
    }
}

[kralli]

interface Foo2 {
    var prop1: String
    var prop2: String

    class Helper(override var prop1: String, override var prop2: String) : Foo2

    companion object {
        operator fun invoke(prop1: String, prop2: String) = Helper(prop1, prop2)
    }
}

[spand]

Still four counts of prop1

[kralli]

Well, you can’t define a function without it’s parameters, but the invoke operator function is the only way to call Foo2 directly.

[spand]

Yeah I thought so too. Except the error message sort of hinted of the possibility of a an operator val.

[Ruckus]

No, it isn't. It knows that operator can't be used there, and to make a nice message, it generates a string representation of your use, which in this case happens to be on a "member property without backing field or delegate".