{"id":25718,"date":"2022-12-13T00:40:29","date_gmt":"2022-12-12T19:10:29","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-convert-character-a-to-b-if-i-add-1-closed\/"},"modified":"2022-12-13T00:40:29","modified_gmt":"2022-12-12T19:10:29","slug":"solved-convert-character-a-to-b-if-i-add-1-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-convert-character-a-to-b-if-i-add-1-closed\/","title":{"rendered":"[Solved] Convert character ‘A’ to ‘B’ if I add 1 [closed]"},"content":{"rendered":"

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Furthering Sergio’s answer, you can isolate the letter-ordinals and use modulo.<\/p>\n

class String\n  def plus n\n    case self\n    when ('a'..'z')\n      (((ord - 97 + n) % 26) + 97).chr            #ord means self.ord\n    when ('A'..'Z')\n      (((ord - 65 + n) % 26) + 65).chr\n    else\n      raise \"single-letters only\"\n    end\n  end\nend\n\n'a'.plus 2 #=> 'c'\n'z'.plus 1 #=> 'a'\n'A'.plus 0 #=> 'A'\n'Z'.plus 9 #=> 'I'\n\"https:\/\/stackoverflow.com\/\".plus 1 #=> test.rb:9:in `plus': single-letters only (RuntimeError)\n<\/code><\/pre>\n
Your question is unclear, but as alluded to in the comments by Stefan, you may want 'A' + 1 #=> 'B'<\/code> notation. Ruby does allow an overwrite of String#+<\/code>. But don’t do this, this is just to show you that you can:<\/p>\n
class String\n  def +(n)\n   #code as above...\n  end\nend\n<\/code><\/pre>\n
then via syntactic sugar, you can do:<\/p>\n
'a' + 2 #=> 'c'\n'z' + 1 #=> 'a'\n'A' + 0 #=> 'A'\n'Z' + 9 #=> 'I'\n\"https:\/\/stackoverflow.com\/\" + 1 #=> test.rb:9:in `plus': single-letters only (RuntimeError)\n<\/code><\/pre>\n<\/p><\/div>\n
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            2<\/span> <\/p><\/div>\n<\/div>\n
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solved Convert character ‘A’ to ‘B’ if I add 1 [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Furthering Sergio’s answer, you can isolate the letter-ordinals and use modulo. class String def plus n case self when (‘a’..’z’) (((ord – 97 + n) % 26) + 97).chr #ord means self.ord when (‘A’..’Z’) (((ord – 65 + n) % 26) + 65).chr else raise “single-letters only” end end end ‘a’.plus 2 #=> ‘c’ … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[455],"class_list":["post-25718","post","type-post","status-publish","format-standard","hentry","category-solved","tag-ruby"],"yoast_head":"\n[Solved] Convert character 'A' to 'B' if I add 1 [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-convert-character-a-to-b-if-i-add-1-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Convert character 'A' to 'B' if I add 1 [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Furthering Sergio’s answer, you can isolate the letter-ordinals and use modulo. class String def plus n case self when ('a'..'z') (((ord - 97 + n) % 26) + 97).chr #ord means self.ord when ('A'..'Z') (((ord - 65 + n) % 26) + 65).chr else raise "single-letters only" end end end 'a'.plus 2 #=> 'c' ... 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