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Ruby

Shawn Vogler
Shawn Vogler
564 Points
Posted by Shawn Vogler
Shawn Vogler
564 Points

How do I return a remainder inside a method?

"Inside the "mod" method, write the necessary code to return the remainder of "a" when divided by "b." Hint: remember the "%" operator!"

So here's what I've got so far:

def mod (a, b)
  puts a / b
  return a%, b%
end

mod (10%), (2%)

It's been so long using the "%" operator. Also, I'm pretty confused about the "return" method.

Shawn Vogler
Shawn Vogler
564 Points
Shawn Vogler
Shawn Vogler
564 Points

Here's what I've updated my code to look like: 

def mod (a, b)
  puts "#{a} divided by #{b} = "
  puts a / b
  puts "The remainder of #{a} divided by #{b} is : "
  puts a % b
end

mod (10), (7)

When I look at the preview it is showing this: "10 divided by 7 = 1 The remainder of 10 divided by 7 is : 3" It's outputting the remainder obviously? What am I missing here?

2 Answers

Shawn Vogler
Shawn Vogler
564 Points
Shawn Vogler
Shawn Vogler
564 Points

I feel like a dummy. All I had to was change it from puts to return. It looks like this.

def mod (a, b)
  puts "#{a} divided by #{b} = "
  puts a / b
  puts "The remainder of #{a} divided by #{b} is : "
  return a % b
end

mod (10), (7)
Vlad Filiucov
Vlad Filiucov
10,665 Points
Vlad Filiucov
Vlad Filiucov
10,665 Points

Hi,

You wrote to much. One line is enough. You don't need return key word in ruby. Last thing method does will be returned so you can get rid of it. You can get rid of puts as well for the same reason as above. And in your method you are trying to display same string 3 times. And also when you are calling your mod method you should pass all arguments inside the brackets, and separated by comma. Also don't leave spaces between method name and argument bracket. So at the end of the day it should look something like this 

def mod(a, b) 
  "The remainder of #{a} divided by #{b} is : #{a%b}"
end

mod(10, 7)