问题:

I came with below solution but I believe that must be nicer one out there ...

array = [ 'first','middle','last']

index = array.length
array.length.times { index -= 1; puts array[index]}

回答1:

array.reverse.each { |x| puts x }

回答2:

Ruby is smart

a = [ "a", "b", "c" ]
a.reverse_each {|x| print x, " " }

回答3:

You can even use a for loop

array = [ 'first','middle','last']
for each in array.reverse do
   print array
end

will print

last
middle
first

回答4:

In case you want to iterate through a range in reverse then use:

(0..5).reverse_each do |i|
  # do something
end

回答5:

If you want to achieve the same without using reverse [Sometimes this question comes in interviews]. We need to use basic logic.

array can be accessed through index
set the index to length of array and then decrements by 1 until index reaches 0

output to screen or a new array or use the loop to perform any logic.

    def reverseArray(input)
      output = []
      index = input.length - 1 #since 0 based index and iterating from 
      last to first

      loop do
        output << input[index]
        index -= 1
        break if index < 0
      end

      output
    end

    array = ["first","middle","last"]

    reverseArray array #outputs: ["last","middle","first"]