Why do newlines within parenthesis change arithmetic results?

Question

Why do the following expressions resolve the way they do? Brackets should have higher precedence than newlines, shouldn't they?

3 - ( 1 + 1 )
# => 1

3 - ( 1
     + 1 )
# => 2

Omitting the plus also lets the expression evaluate to 2:

3 - ( 1
      1 )
# => 2

If I declare as a continuous newline (escaped) or move the plus to the first line, desired behavior is achieved:

3 - ( 1 \
     + 1 )
# => 1

3 - ( 1 +
      1 )
# => 1

Answer 1

It is because Ruby recognizes a new line as an end of an expression unless the expression is incomplete. For example,

(1
+ 1)

is the same as

(1;
+1)

which is the same as +1 since the last expression within the parentheses is returned. And that is further the same as 1.

When you have a + at the end of a line, the expression is incomplete, and hence continues to the next line. That makes:

3 - ( 1 +
      1 )

to be interpreted as 3 - (1 + 1).

Answer 2

If you have code in the brackets then every line will be threat as separated code line of code if you won't end it with \ or start new one with math operator.

So in your example:

def plus_minus_if_you_understand_this_the_problem_is_solved_i_guess 
    3 - (1
         1 )
end

Means I have number 3 and want to subtract expression in brackets. In the brackets I have line #1 number 1 and in #2 line number 1 again and as it's last line of expression it's retuned by Ruby (like in def last item before end is returned. So:

( 3   # doing nothing
  1 ) # returns 1

Also take a look bellow. Again, this part of code returns 2 as it is the last item in brackets:

( puts "hello!"
  2 ) => 2