はじめに 前回の記事 #1 で「Scannerは読み取り係」だと分かりました。 今回は いちばんよく使う基本技、nextLine() で「質問→1行読む→すぐ返事」を作ります。シンプルだけど、実務でもこれが土台です。 🎓 今日やること（ゴール） キーボードから 1行...

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Apart from reading files, Scanner can also read user input from Console in Java. Just like in the case of reading files, we have provided File as a source for scanning, We need to provide System.in as a source to scan for user input in Console. Once you created and initialized java.util.Scanner, you can use its various read method to read input from users. If you want to read String, you can use nextLine(), if you want to read integer numbers, you can use nextInt().  --- Java Interview questions and tutorials

### Godot Version 4.4 ### Question Hello, my function’s not returning an array while within my loop. Am I doing something wrong here var returnArray: Array = createTab(nextLine, nextText, choiceName, indent) full code bellow, im trying to make a dialog dictionary that branches out infinity. is giving the error "Trying to assign value of type ‘Nil’ to a variable of type ‘Array’. It’s recursive so it should be looping all the way through and then bringing the return back right? or is there something else I need to do to make sure that it returns the array to the original call? Thank you! func branchCreator(lineNumber: int, choiceEnum, choiceName, tabs): #WOOOOOO RECURSIVE FUNCTIONS LETS GO var indent = tabs var previousIndent: int = tabs var text: String = choiceName var nextText = allScript[lineNumber + 1] var nextLine = lineNumber + 1 choiceName = {} choiceName[lineNumber] = [text, choiceEnum] if text.contains("->") == true: #gets amount of tabs var regex = RegEx.new() regex.compile("\t") var result = regex.search(text) if result != null: var string = result.get_string() indent = string.length() else: indent = 0 while nextText.contains("\t"): var returnArray: Array = createTab(nextLine, nextText, choiceName, indent) print(str(returnArray)) return choiceName else: return choiceName func createTab(lineNumber: int, text, choiceName, choiceIndent: int): var newLineNumber = lineNumber + 1 var assignedLine = assignLineType(allScript[lineNumber]) var indentComparison = choiceIndent var indent if text.contains("->") == true: #gets amount of tabs and checks if its a choice var regex = RegEx.new() regex.compile("\t") var result = regex.search(text) if result != null: var string = result.get_string() indent = string.length() else: indent = 0 if indent <= choiceIndent: return [lineNumber, text, choiceName] #end and go back to loop if indent > choiceIndent: branchCreator(lineNumber, assignedLine, text, indent) #make a new branch else: #makes the tab and then makes another tab choiceName[lineNumber] = [text, assignedLine] createTab(newLineNumber, allScript[newLineNumber], choiceName, choiceIndent)

YoungFry: > > if text.contains("\t") == true: #gets amount of tabs > var regex = RegEx.new() > regex.compile("\t") > var result = regex.search(text) > print(str(result)) > elif nextCheck.contains("\t") == true: > tabChecker(nextLine, nextCheck, choiceName ) > if indent > 0: > tabChecker(nextLine, nextCheck, choiceName ) > 1. What is the point of this first condition `text.contains("\t") == true` statement block? It’s only doing a RegEx search and then printing the result. 2. I think the second condition `elif nextCheck.contains("\t") == true` is not what you may be intending. Consider the input for line `\t → choice 2`; This ends up not running the `nextCheck` condition block because the first `text.contains("\t")` condition was true, even though the nextCheck line would be returning true. 3. What is the point of `if indent > 0:` conditional? You set `var indent: int = 0` at the top of the function but never modify it. This conditional block never runs. The issues with #2 and #3 seem to explain why you may not be getting down into the next Dictionary tab level.

Scanner sc = new Scanner ( System . in ); What do you do when you want to read an int in Java? You would probably write sc.nextInt() . And what about string? You would say sc.next() or sc.nextLine()...

The nextLine() Issue: Understanding the Scanner Buffer in Java Almost every Java learner encounters this problem at some point. You write a simple program that asks for an age ( int ) and then a name...