Understanding loop iteration in Ruby

Question

I have written this code to determine how many a's are present in a given input within a method:

def count_a(word)
  count = 0
  i = 0
  while i < word.length
    char = word[i]
    
    if char == "a" || char == "A"
        count += 1
    end 
    
    i += 1  
  end

  return count 
end

puts count_a("application")  # => 2
puts count_a("bike")         # => 0
puts count_a("Arthur")       # => 1
puts count_a("Aardvark")     # => 3

Can someone better explain why I need the i += 1 at the bottom of the method? I have added to the count in the conditional above, so I just need some further knowledge on how the iteration becomes necessary outside the conditional and within the method.

Thanks!

The code failed to run without the i += 1

Don't forget char.downcase == 'a' or case char and when /\Aa\z/i which ignores case. — tadman
– tadman, Commented Dec 6, 2022 at 16:08
Try that most basic of debugging techniques, stepping through the program by hand. Write the results of each statement on a piece of paper. Don’t anticipate or intuit the outcomes, just follow the instructions literally, because that’s what the computer is going to do. Do this both with and without the line that has you puzzled. By the way, this is a major incentive to keep methods short and sweet — it’s no fun going through hundreds of lines of code by hand. — pjs
– pjs, Commented Dec 6, 2022 at 16:39
Now that your question has been answered you should consider posting your code at SO's sister-site Code Review. At that forum you can expect to get excellent advice on how to improve your code by using an assortment of Ruby's many built-in methods. Answers there will also help you gain a better understanding of the Ruby way of thinking about coding problems. — Cary Swoveland
– Cary Swoveland, Commented Dec 6, 2022 at 18:45
If you don't ever increment i, the variable would always have the value 0, and the condition for while would always be true, leaving you with an infinite loop. — user1934428
– user1934428, Commented Dec 7, 2022 at 7:46

max · Accepted Answer · 2022-12-06 17:07:51Z

2

Thats a decent first attempt. But its not really Ruby.

Ruby actually has tons of methods that lets you complete this without mutating an external variable:

def count_a(word)
  word.each_char # enumerate over the chars
      .count { |l| ['a', 'A'].include?(l) } # counts the number of elements yielding a true value.
end

In general methods like while, until and loop should mainly be used when you're dealing with input or streams - in other words when you're not looping across an object with a known length. For everything else use the plethory of methods provided by Enumerable and the other core classes.

Can someone better explain why I need the i += 1 at the bottom of the method?

Because otherwise the loop will run forever since i < word.length will always be true. But as a rule of thumb if you ever write i += 1 in Ruby you're doing it wrong.

edited Dec 6, 2022 at 17:07

answered Dec 6, 2022 at 16:47

max

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1 Comment

steenslag Over a year ago

word.count("Aa")

David Grayson · Accepted Answer · 2022-12-06 17:34:37Z

1

The line i += 1 adds one to i. If you remove it, then i will be stuck at 0 forever because you never change it. Therefore, your loop will never terminate (in the case that word.length is positive).

answered Dec 6, 2022 at 17:34

David Grayson

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Comments

Chris · Accepted Answer · 2022-12-06 16:14:50Z

0

Any loop needs a test and an update. If the variables in your test never change, why would the loop ever end?

This method could be written far more idiomatically as follows. By downcasing the string first we don't have to check for 'a' and 'A'.

def count_a(word)
  word.downcase.count('a') 
end

As an intermediate solution, you can use #each_char to iterate over each character in the string.

def count_a(word)
  count = 0
  word.each_char do |ch|
    count += 1 if ch == 'a' || ch == 'A'
  end
  count

Be aware that a method returns its last expression, an explicit return is only needed to break control flow and return early from a method.

edited Dec 6, 2022 at 16:14

answered Dec 6, 2022 at 16:07

Chris

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1 Comment

steenslag Over a year ago

word.count("Aa")

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