Methods in Ruby implicitly returns their last line unless it’s preceeded by an explicit
def random_method(str) str = 'new value' end new_str = random_method('first value') puts '---------' puts new_str
On the example above,
str = 'new value' is implicitly returned.
This method would actually anger Rubocop because we are assigning a string to a variable which will be returned anyway. We could (and should) directly return the string.
2:3: W: Lint/UselessAssignment: Useless assignment to variable - str. (https://github.com/bbatsov/ruby-style-guide#underscore-unused-vars) str = 'new value' ^^^
The implicit return value is important because if not well understood it can be the cause of a lot of problems down the road. Let’s say for example we want to output our new value:
def random_method(str) str = 'new value' puts str end new_str = random_method('first value') puts '---------' puts new_str
One would think the code above outputs
new_value two times. But it doesn’t because the method
random_method’s last line is
puts str. The method
#puts always returns
nil, which is what
random_ method returns as well.
It is possible to explicitly return a value using the
def random_method(str) return 'new value' str = 'newiest value' end new_str = random_method('first value') puts '---------' puts new_str
'new value' is returned and the code after is not executed. It’s aid to b unreachable.
This is most well illustrated by the `#select` method:
(1..10).to_a.select do |num| num.odd? end # => [1, 3, 5, 7, 9] (1..10).to_a.select do |num| puts num.odd? end # => 
Blocks also implicitely return the last statement (and the only one here).
#select check the return result of each block and add it to a new array if the result evaluates to
true. For the second range, the
#puts method always return
nil which evaluates to
false, which is why the new array is empty.