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Ruby Ruby Basics (Retired) Ruby Methods Method Returns: Part 2

Posted by andynguyen9
andynguyen9
1,629 Points

Why these codes are not correct? def mod(a, b) c = a % b puts "The remainder of a divided by b is #{c}" end

I tested these codes in irb and it works but not in Workspace.

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

method.rb 
def mod(a, b)
  c = a%b
  puts "The remainder of a divided by b is #{c}"
end

3 Answers

Steve Hunter
Steve Hunter
57,712 Points

Hi Andy,

Interpolate a and b too so the values appear in your string. Do the same as you did with c. Note, you can interpolate an expression, so you didn't need to create c, interpolating #{a % b} is fine.

Also, you want to return the string, not puts it.

Make sense?

Steve.

andynguyen9
andynguyen9
1,629 Points
andynguyen9
andynguyen9
1,629 Points

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

Bummer in my console. Is this code correct?

Steve Hunter
Steve Hunter
57,712 Points

Not quite right there; delete puts and type return instead.

Steve Hunter
Steve Hunter
57,712 Points

Strictly, you don't need the return keyword at all as Ruby returns the last line evaluated in every method. But sometimes it's just clearer to use the return keyword as it makes it totally obvious what you intend to happen.

So, you could just have:

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

But it is a little clearer like this:

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

Steve.

andynguyen9
andynguyen9
1,629 Points
andynguyen9
andynguyen9
1,629 Points 
def mod(a,b)
  return "The remainder of #{a} divided by #{b} is #{a%b}"
end

Thank you Steve, I got it.

I love Treehouse!

Steve Hunter
Steve Hunter
57,712 Points

:+1: :smile: