zipcode 07020 zipcode 07020

coding for food

• [Ruby] Reverse

  coding for food 2022. 7. 23. 13:00

  Write a method reverse(word) that takes in a string word and returns the word with its letters in reverse order. my code def reverse(word) rev = "" i = word.length while i > 0 i -= 1 rev += word[i] end return rev end puts reverse("cat") # => "tac" solution def reverse(word) reversed = "" i = 0 while i < word.length char = word[i] reversed = char + reversed i += 1 end return reversed end puts rev..

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• [Ruby] Sum Nums

  coding for food 2022. 7. 23. 03:32

  my code def sum_nums(max) i = 0 sum = 0 while i 10, because 1 + 2 + 3 + 4 = 10 puts sum_nums(5) # => 15 solution def sum_nums(max) sum = 0 i = 1 while i 10 puts sum_nums(5) # => 15 1. sum_nums(4), 구해야 하는 답이 1+2+3+4 = 10 인데 처음 시작이 1부터 max 4 이기 때문에 i = 1 로 시작하고 while에서의 조건 i 는 max 보다 작거나 같아야 한다. I wanna take i and add it to my sum, but..

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• [Ruby] Sum Nums

  coding for food 2022. 7. 22. 14:16

  my code def sum_nums(max) i = 0 sum = 0 while i 10, because 1 + 2 + 3 + 4 = 10 puts sum_nums(5) # => 15 solution def sum_nums(max) sum = 0 i = 1 while i 10 puts sum_nums(5) # => 15

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• [Ruby] Count E

  coding for food 2022. 7. 22. 11:49

  Count E Write a method count_e(word) that takes in a string word and returns the number of e's in the word def count_e(word) count = 0 # use count to track the number of e's i = 0 # use i to iterate thru the word while i 1 puts count_e("excellent") # => 3

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• [Ruby] break, next

  coding for food 2022. 7. 21. 13:50

  # break - immediately exist the loop # next - skips to the next iteration i = 1 while 1

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• [Ruby] String Iteration

  coding for food 2022. 7. 21. 13:17

  def printChars(str) i = 0 while i < str.length puts str[i] i += 1 end end printChars("pizza") p i z z a

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• [Ruby] While Loop

  coding for food 2022. 7. 21. 09:53

  def repeatHello () counter = 1 while counter

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• [Ruby] Conditionals exercises

  coding for food 2022. 7. 21. 08:21

  my code 1 def either_only(number) if number % 15 == 0 return false elsif number % 3 == 0 return true elsif number % 5 == 0 return true else return false end end puts either_only(9) # => true puts either_only(20) # => true puts either_only(7) # => false puts either_only(15) # => false puts either_only(30) # => false solution 2 def either_only(number) if (number % 3 == 0 || number % 5 == 0) && !(n..

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