Hello everyone, I was playing around with val func...

# getting-started

[Matias Reparaz]

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]

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() })

[Matias Reparaz]

oka! perfect! that makes sense! Thank you!

[Matias Reparaz]

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

[Matias Reparaz]

and I can add another option to the list f("hello, world").invoke {it.uppercase()}.also { println("f: $it") }

[dmitriy.novozhilov]

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

[Matias Reparaz]

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]

f() == f.invoke() (if f is a val, var, object)
f() {...} == f({...}) == f {...} empty parens before lambda can be present or omitted

[ephemient]

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

[Matias Reparaz]

perfect, crystal clear! thanks!