`fun foo() = println("No receiver") fun Int.foo() ...

# announcements

[Astronaut4449]

fun foo() = println("No receiver")
fun Int.foo() = println("My receiver is $this")

with(42) { foo() } // will print 'My receiver is 42' but I want 'No receiver'

Is it somehow possible to call the function without receiver, although one is given by the current scope?

[bezrukov]

if foo() is not top level fun, then you can specify it via `this`:

with(42) { [e-mail]() }

[Vampire]

Depends on more context.
For example you can qualify it if it is a top-level function like

package bar

fun foo() = println("No receiver")
fun Int.foo() = println("My receiver is $this")

fun main(){
    with(42) { bar.foo() }
}

Or if the foo is a local function it will also win:

fun Int.foo() = println("My receiver is $this")

fun main(){
    fun foo() = println("No receiver")
    with(42) { foo() }
}

A combination of the two:

package bar

fun foo() = println("No receiver")
fun Int.foo() = println("My receiver is $this")

fun main(){
    fun foo() = bar.foo()
    with(42) { foo() }
}

Or you can qualify the this reference if possible like

class Foo {
    fun foo() = println("No receiver")
    fun Int.foo() = println("My receiver is $this")

    fun main(){
        with(42) { [e-mail]() }
    }    
}
fun main() = Foo().main()

[Astronaut4449]

fun foo() = println("No receiver")
fun Int.foo() = println("My receiver is $this")

fun main() {
    with(42) { foo() } // will print 'My receiver is 42' but I want 'No receiver'
}

Sorry for being so scarce on the context.

Both functions are top level declarations.

[Vampire]

Anything wrong with the options I gave you?

[Astronaut4449]

Thank you. I guess I need to use the package name. Unfortunately when I tested this I used no package 🤣