Sort method in Ruby - arrays

Question

I wrote method, to sort array of user inputs(integers) from low to high and reverse. Code looks like:
def alphabetize(arr, rev=false)
arr.sort!
if rev==true
arr.reverse!
end
return arr
end
puts "enter your numbers with 'space' between them"
text = gets.chomp
numbers = text.split(" ")
numbers.each do |element|
element.to_i
end
puts alphabetize(numbers)
First time i tried only numbers from 0 to 9, and method had worked correct. But then i tried to input numbers like 1112, 11, 22 and after sorting procedure, i had got result like this "11 1112 22". After this, i tried to change code this way:
def alphabetize(arr, rev=false)
arr.sort!{|a, b| a.to_i <=> b.to_i}
if rev==true
arr.reverse!
end
return arr
end
puts "enter your numbers with 'space' between them"
text = gets.chomp
numbers = text.split(" ")
puts alphabetize(numbers)
And this way my code works correct and sorting input of 11, 1112, 22 produce the result that i expect: "11, 22, 1112". Looks like i have solved the problem, but i really want to understand the difference between first version and second version of code. Why method sort works in different ways?

Answer 1

each does not replace the elements in an array - it simply returns the array itself (without change).
You might prefer to use map:
numbers = numbers.map do |element|
element.to_i
end
Or, better yet, use map!, which actually changes the array itself, rather than return the changed array:
numbers.map! do |element|
element.to_i
end
You can also use the following shorthand:
numbers.map! &:to_i

Answer 2

Looks like in the first place you still comparing strings instead of integers.
In fact, in the first place you don't convert elements to integers:
numbers.each do |element|
element.to_i
end
The script below don't replace strings with integers in array.