CODEWITHSUNDEEP
ruby
Dylan T
Dylan T

Reputation: 139

Having trouble figuring out a ruby math error. advice?

Multiplication is the only operator that works. every other operator

still multiplies for some reason. for example, 5+5=25, 6-2=12, 6/2=12.

It looks as though all my operators are correct in the code, so I am

not quite sure why this is happening. 

puts "Enter your first Number: "
first_num = gets.to_i

puts "Enter your modifier"
modifier = gets

puts "Enter second number"
second_num = gets.to_i

def add(first_num, second_num)
  return first_num + second_num
end

def subtract(first_num, second_num)
  return first_num - second_num
end

def multiply(first_num, second_num)
  return first_num * second_num
end

def divide(first_num, second_num)
  return first_num / second_num
end

case modifier
when +
  final_num = add(first_num, second_num)
when -
  final_num = subtract(first_num, second_num)
when *
  final_num = multiply(first_num, second_num)
when %
  final_num = divide(first_num, second_num)
end

puts final_num

Upvotes: 1

Views: 113

Answers (2)

RainbowNinja
RainbowNinja

Reputation: 51

Put " " around your operators in your case statements.So case modifier when "+" etc .gets saves things as a string, so your modifier is saved as a string.

Upvotes: 0

spickermann
spickermann

Reputation: 106802

gets returns strings. More specifically gets returns the string entered by the user including the line break which is appended when the user presses ENTER.

Therefore, it makes sense to first remove the new line from the user input with gets.chomp. And second, compare modifier with strings containing operators in the case block:

puts "Enter your modifier"
modifier = gets.chomp

# ...

case modifier
when '+'
  final_num = add(first_num, second_num)
when '-'
  final_num = subtract(first_num, second_num)
when '*'
  final_num = multiply(first_num, second_num)
when '%'
  final_num = divide(first_num, second_num) 
end

Or insted of the case block a refactored version using a hash and send:

OPS = { '+' => 'add', '-' => 'substract', '*' => 'multiply', '%' => 'divide' }
final_num = send(OPS[modifier], first_num, second_num)

Actually, I am surprised that Ruby doesn't raise a syntax error.

Upvotes: 3

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