What is the logic behind this result?

Question

def foo(_, _='override')
  _
end

p foo("bye bye")
p foo("hello", "world")

Output:

"override"
"hello"

I could understand if the result was:

"override"
"world"

or even:

"bye bye"
"hello"

But the result I'm getting causes me confusion.

Answer 1

Default arguments are evaluated earlier in time than regular arguments if an argument is passed for it, otherwise they are evaluated last. Almost certain but unsure how to prove it.

Meaning in this example:

at time 0 call p foo("hello", "world")

at time 1 _ = 'override'

at time 2 _ = "world"

at time 3 _ = "hello"

at time 4 variables are printed and you see "hello"

EDIT here is some evidence:

def foo(_, _='override',_)
  _
end

p foo("bye bye","goodbye")
p foo("hello", "world", "three")

prints

"override"
"three"

Answer 2

Ican't find better explanation than this one

ruby magic underscore

The reason for this is found in Ruby’s parser, in shadowing_lvar_gen. All the normal checks for duplication are skipped iff the variable name consists of exactly one underscore.