Asking out of lack of experience with R: how does such invocation handle case when `x` is defined with a different value at call site?
In pseudocode:
f =
let x = 1 in # inner vars for f go here
arg -> arg + 1 # function logic goes here
# example one: no external value
f (x+1) # produces 3 (arg := (x+1) = 2; return arg +1)
# example two: x is defined in the outer scope
let x = 4 in
f (x+2) # produces 5 (arg := 4; return arg + 1)? Or 3 if inner x wins as in example one?
In pseudocode: