Hello everyone I was playing around with val functions and I kotlinlang #getting-started

Hello everyone, I was playing around with val functions and I ran into this problem ... could someone help me understand why the e option doesn’t work (which would seem the most reasonable) and why all the above works? https://pl.kotl.in/YdmU2qEUy

dmitriy.novozhilov

06/25/2021, 1:23 PM

When you use trailing lambda syntax it is just syntax sugar for such call:

foo(a, b, c) { bar() } 
foo(a, b, c, { bar() })

So when you call

f("hello, world") { it.toUpperCase() }

in your example is is desugared to

f.invoke("hello, world", { it.toUpperCase() })

, which is incorrect, because it should be

f.invoke("").invoke({ it.toUpperCase() })

☝️ 1

Matias Reparaz

06/25/2021, 2:44 PM

oka! perfect! that makes sense! Thank you!

with that in mind I understand all the examples except the A.. how does A works?

and I can add another option to the list

f("hello, world").invoke {it.uppercase()}.also { println("f: $it") }

dmitriy.novozhilov

06/25/2021, 5:17 PM

f("hello, world")() { it.uppercase() }
// same as
val x = f("hello, world")
x() { it.uppercase() }

Matias Reparaz

06/25/2021, 5:24 PM

that is example D, and there is no need to use the parentheses... even more, why I can add the parentheses if the

val a

has type

((String) -> String) -> String

ephemient

06/25/2021, 5:56 PM

f() == f.invoke()

(if

is a val, var, object)

f() {...} == f({...})  == f {...}

empty parens before lambda can be present or omitted

in that case,

f1 { it.uppercase() }
f1() { it.uppercase() }
f1({ it.uppercase() })
f1.invoke { it.uppercase() }
f1.invoke() { it.uppercase() }
f1.invoke({ it.uppercase() })

are all equivalent

👏 1

Matias Reparaz

06/25/2021, 6:05 PM

perfect, crystal clear! thanks!

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