Explanation of the FizzBuzz problem

FizzBuzz problem commentary

I also summarized Expression used for FizzBuzz problem.

** Step 1 ** Output numbers from 1 to 100 to the terminal

** Step 2 ** When it is "multiple of 3", Fizz is a character string instead of a number. Similarly for "multiples of 5", Buzz

** Step 3 ** Output as FizzBuzz when it is a multiple of 3 and 5, which is a multiple of 15.

Program template

Template

def fizz_buzz 
#Contents of processing
end

fizz_buzz

A program to output numbers from 1 to 100 to the terminal

step 1 Output numbers from 1 to 100 to the terminal

Output numbers from 1 to 100 to the terminal

def fizz_buzz
  num = 1
# num =Contents of processing starting from 1
  while num <= 100 do #As long as the boolean value is true, the following processing continues
    puts num #Output numerical value

    num = num + 1 #To boolean value+1
  end
#Contents of processing
end

fizz_buzz

When it is "multiple of 3", it outputs Fizz as a character string, and when it is "multiple of 5", it outputs Buzz.

** Step 2 ** When it is a multiple of 3, use a character string instead of a number. Buzz` for" multiples of 5 "

When it is "multiple of 3", it is output as Fizz as a character string, and when it is "multiple of 5", it is output as Buzz.

def fizz_buzz
  num = 1
# num =Contents of processing starting from 1
  while num <= 100 do #As long as the boolean value is true, the following processing continues
    if num % 3 == 0 #When the remainder divided by 3 is 0
      puts "Fizz"
    elsif num % 5 == 0 #When the remainder after dividing by 5 is 0
      puts "Buzz"
    else #At other times
      puts num #Numerical values ​​that do not correspond to the above multiples of 3 and 5 are output as they are.
    end
    num = num + 1 #To boolean value+1
  end
#Contents of processing
end

fizz_buzz

Add a conditional expression for "multiples of 15 (multiples of 3 and 5)"

** Step 3 ** FizzBuzz and output when it is a multiple of 3 and 5, which is a multiple of 15.

Added conditional expression for "multiples of 15 (multiples of 3 and 5)"

def fizz_buzz
  num = 1
# num =Contents of processing starting from 1
  while (num <= 100) do #As long as the boolean value is true, the following processing continues
    if num % 15 == 0 #When it is a multiple of 15
      puts "FizzBuzz"
    elsif (num % 3) == 0 #When the remainder divided by 3 is 0
      puts "Fizz"
    elsif (num % 5) == 0 #When the remainder after dividing by 5 is 0
      puts "Buzz"
    else #At other times
      puts num #Numerical values ​​that do not correspond to multiples of 3, 5, and 15 above are output as they are.
    end
    num = num + 1 #To boolean value+1
  end
#Contents of processing
end

fizz_buzz

Notes If you add it below, the above conditional expression "is it a multiple of 3" or "is it a multiple of 5" becomes true and is displayed as Fizz or Buzz.

From the above points, add a condition to output FizzBuzz when it is a "multiple of 15" at the beginning of the if statement.

• Alternative conditional expression "when it is a multiple of 15"
• The ones mentioned above are num% 15 == 0
• Can be replaced with num% 3 == 0 && num% 5 == 0.