Spoiler alert: I am a true novice. Tasked with figuring out fizz buzz in ruby for a class and while I have found more than a few versions of code that solve the problem, my understanding is so rudimentary that I cannot figure out how these examples truly work.
First question(refer to spoiler alert if you laugh out loud at this): How do i print out numbers one through 100 in Ruby?
Second question: can 'if else" be used to solve this? My failed code is below(attachment has screen shot):
puts('Lets play fizzbuzz')
print('enter a number: ')
number = gets()
puts(number)
if number == % 3
puts ('fizz')
elsif number == % 5
puts ('buzz')
elsif number == %15
puts ('fizzbuzz')
end
Thanks,
Thats ok being a novice, we all have to start somewhere right? Ruby is lovely as it get us to use blocks all the time, so to count to 100 you can use several methods on fixnum, look at the docs for more. Here is one example which might help you;
1.upto 100 do |number|
puts number
end
For your second question maybe take a quick look at the small implementation i whipped up for you, it hopefully might help you understand this problem:
1.upto 100 do |i|
string = ""
string += "Fizz" if i % 3 == 0
string += "Buzz" if i % 5 == 0
puts "#{i} = #{string}"
end
First question: this problem has several solutions. For example,
10.times { |i| puts i+1 }
For true novice: https://github.com/bbatsov/ruby-style-guide
another method that can be helpful :
puts (1..100).map {|i|
f = i % 3 == 0 ? 'Fizz' : nil
b = i % 5 == 0 ? 'Buzz' : nil
f || b ? "#{ f }#{ b }" : i
}
In Regards to your failed code, your conditional statements should be like this:
if number % 3 == 0
puts "Fizz"
end
if number % 5 == 0
puts "Buzz"
end
You don't want the last elsif statement because it will never get executed (if a number is not divisible by 3 or divisible by 5, then it is certainly not divisible by 15) Adjust for this by changing the second elsif to simply and if and if the number is divisble by 5 and not by 3, then Fizz will not be outputted but Buzz Will be
I'm just showing you how to correct your code, but as others have pointed out, there are far more elegant solutions in Ruby.
Not the most beautiful way to write it but good for beginners and for readability.
def fizzbuzz(n)
(1..n).each do |i|
if i % 3 == 0 && i % 5 == 0
puts 'fizzbuzz'
elsif i % 3 == 0
puts 'fizz'
elsif i % 5 == 0
puts 'buzz'
else
puts i
end
end
end
fizzbuzz(100)
1.upto(100).each do |x| # Question #1 The 'upto' method here takes is
# what you would use to count in a range.
if (x % 3 == 0) && (x % 5 == 0)
puts " Fizzbuzz"
elsif x % 3 == 0
puts " Fizz"
elsif x % 5 == 0
puts " Buzz"
else
puts x
end
end
Question #2 Yes you can but I would look for a more elegant way to write this as a part of a definition like
def fizzbuzz(last_number)
1.upto(last_number).each do |x|
if (x % 3 == 0) && (x % 5 == 0)
puts " Fizzbuzz"
elsif x % 3 == 0
puts " Fizz"
elsif x % 5 == 0
puts " Buzz"
else
puts x
end
end
end
This is the answer that helped me to understand that no variables are being created with the .each method. Sorry about my indenting. Still learning how to use Stackoverflow text editing.
Here is my most "idiomatic ruby" solution:
class FizzBuzz
def perform
iterate_to(100) do |num,out|
out += "Fizz" if num.divisable_by?(3)
out += "Buzz" if num.divisable_by?(5)
out || num
end
end
def iterate_to(max)
(1..max).each do |num|
puts yield num,nil
end
end
end
class Fixnum
def divisable_by?(num)
self % num == 0
end
end
class NilClass
def +(other)
other
end
end
FizzBuzz.new.perform
And it works:
First question:
for i in 1..100
puts i
end
来源：https://stackoverflow.com/questions/21666030/fizz-buzz-in-ruby-for-dummies