Meget simpelt Hangman-spil

Følger denne kode standardkonventioner, ikke overflødig? Hvordan kan jeg gøre det mere forståeligt for andre mennesker?

Det er en øvelse fra Java-introduktionsbogen:

Skriv en hangman-spil, der tilfældigt genererer et ord og beder brugeren om at gætte et bogstav ad gangen. Hvert bogstav i ordet vises som en stjerne. Når brugeren foretager et korrekt gæt, vises det aktuelle bogstav. Når brugeren er færdig med et ord, vis antallet af savner.

Mit program:

import java.util.Arrays; import java.util.Scanner; public class Hangman{ public static void main(String[] args) { String[] words = {"writer", "that", "program"}; // Pick random index of words array int randomWordNumber = (int) (Math.random() * words.length); // Create an array to store already entered letters char[] enteredLetters = new char[words[randomWordNumber].length()]; int triesCount = 0; boolean wordIsGuessed = false; do { // infinitely iterate through cycle as long as enterLetter returns true // if enterLetter returns false that means user guessed all the letters // in the word e. g. no asterisks were printed by printWord switch (enterLetter(words[randomWordNumber], enteredLetters)) { case 0: triesCount++; break; case 1: triesCount++; break; case 2: break; case 3: wordIsGuessed = true; break; } } while (! wordIsGuessed); System.out.println("\nThe word is " + words[randomWordNumber] + " You missed " + (triesCount -findEmptyPosition(enteredLetters)) + " time(s)"); } /* Hint user to enter a guess letter, returns 0 if letter entered is not in the word (counts as try), returns 1 if letter were entered 1st time (counts as try), returns 2 if already guessed letter was REentered, returns 3 if all letters were guessed */ public static int enterLetter(String word, char[] enteredLetters) { System.out.print("(Guess) Enter a letter in word "); // If-clause is true if no asterisks were printed so // word is successfully guessed if (! printWord(word, enteredLetters)) return 3; System.out.print(" > "); Scanner input = new Scanner(System.in); int emptyPosition = findEmptyPosition(enteredLetters); char userInput = input.nextLine().charAt(0); if (inEnteredLetters(userInput, enteredLetters)) { System.out.println(userInput + " is already in the word"); return 2; } else if (word.contains(String.valueOf(userInput))) { enteredLetters[emptyPosition] = userInput; return 1; } else { System.out.println(userInput + " is not in the word"); return 0; } } /* Print word with asterisks for hidden letters, returns true if asterisks were printed, otherwise return false */ public static boolean printWord(String word, char[] enteredLetters) { // Iterate through all letters in word boolean asteriskPrinted = false; for (int i = 0; i < word.length(); i++) { char letter = word.charAt(i); // Check if letter already have been entered bu user before if (inEnteredLetters(letter, enteredLetters)) System.out.print(letter); // If yes - print it else { System.out.print("*"); asteriskPrinted = true; } } return asteriskPrinted; } /* Check if letter is in enteredLetters array */ public static boolean inEnteredLetters(char letter, char[] enteredLetters) { return new String(enteredLetters).contains(String.valueOf(letter)); } /* Find first empty position in array of entered letters (one with code \u0000) */ public static int findEmptyPosition(char[] enteredLetters) { int i = 0; while (enteredLetters[i] != "\u0000") i++; return i; } } 

Log af sit arbejde:

 (Guess) Enter a letter in word **** > a (Guess) Enter a letter in word **a* > t (Guess) Enter a letter in word t*at > q q is not in the word (Guess) Enter a letter in word t*at > t t is already in the word (Guess) Enter a letter in word t*at > b b is not in the word (Guess) Enter a letter in word t*at > h (Guess) Enter a letter in word that The word is that You missed 2 time(s)  

Kommentarer

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• din kode er for kompliceret, du skal forklare, hvad der foregår, da det er svært for en studerende at læse og vide nøjagtigt hvad ' sker. Tak.