CODEWITHSUNDEEP
rubyvariable-assignment
alistair_Welles
alistair_Welles

Reputation: 69

Array element access in Ruby

I was trying to create a Table of Contents in the command line. Each element in the array is a string variable. 

arr = [chap1, chap1_page, chap2, chap2_page, chap3, chap3_page]
x = 0

until x == arr.length
if ((arr[x] != 0 ))
    puts arr[x].ljust(line/2) + arr[x += 1].rjust(line/2)       
end 
x += 1
end

I was wondering if someone could explain the second half of the puts statement. I'm not sure why arr[x+=1] works but arr[x+1] does not. As far as I know, they are the same, are they not?

Upvotes: 1

Views: 142

Answers (2)

Keith Bennett
Keith Bennett

Reputation: 4950

When dealing with Enumerables like arrays and hashes, it's useful to search the documentation to see if there's something there that will make your code higher level and more expressive. In this case, you can use each_cons to give you the pairs so you don't need to use array indexes at all:

2.3.0 :004 > [1,2,3,4].each_cons(2).to_a
 => [[1, 2], [2, 3], [3, 4]]

Also, rather than using if statements, it's better IMO to use select and reject.

Also, intermediate local variables can make your code more readable.

Using these ideas, your code could look something like this:

array = [chap1, chap1_page, chap2, chap2_page, chap3, chap3_page]

width = line / 2

array.each_cons(2).reject { |x,y| x == 0 }.each do |left, right|
  puts left.ljust(width) + right.ljust(width)
end

(I haven't tested this code, but it shows the general idea.)

You could break down those enumerable calls and assign intermediate values to local variables if that makes it clearer for you:

array = [chap1, chap1_page, chap2, chap2_page, chap3, chap3_page]
width = line / 2

pairs = array.each_cons(2)
nonzero_pairs = pairs.reject { |x,y| x == 0 }
nonzero_pairs.each do |left, right|
  puts left.ljust(width) + right.rjust(width)
end

Upvotes: 1

sawa
sawa

Reputation: 168071

x + 1 returns that value and has no side effect (does not change the reference of x). x += 1 reassigns x and returns that value.

Upvotes: 0

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