我想創(chuàng)建簡單的控制臺僚機游戲。我的錯誤是，如果我嘗試在單詞中獲取 2 個相同字符的 pos，我只會得到一個，而另一個被跳過。例如西紅柿。控制臺輸出：番茄_ 奧馬特 _ _我知道我沒有使用 live 沒有時間我做它層。類程序{ static string[] word = { "Pineapple", "Apple" , "Tomatoe" , "Pizza"}; static int wordIndex = 0; static char[] randomWord; static bool guessing = true; public static void Main(string[] args) { int lives = 3; Console.OutputEncoding = Encoding.UTF8; Console.InputEncoding = Encoding.UTF8; Random r = new Random(); wordIndex = r.Next(word.Length); randomWord = word[wordIndex].ToLower().ToCharArray(); char[] randomWordcensored = new char[randomWord.Length]; for (int i = 0; i < randomWord.Length; i++) { randomWordcensored[i] = '_'; } Console.WriteLine("Hello"); foreach (var item in randomWordcensored) { Console.Write(item + " "); } Console.WriteLine(); Console.WriteLine("Please Enter character:"); while (guessing = true) { int g = 0; char userinput; bool security = char.TryParse(Console.ReadLine() ,out userinput); if (security == true) { if (randomWord.Contains(userinput)) { //help needed g = (word[wordIndex].ToString().IndexOf(userinput) == -1 ? 0 : word[wordIndex].ToString().IndexOf(userinput)); randomWordcensored[g] = userinput; Console.WriteLine("Good :) " + g); foreach (var item in randomWordcensored) { Console.Write(item + " "); } } else { lives--; Console.WriteLine("Wrong!\n-Lives:" + lives); } } else { Console.WriteLine("Enter only one charracter!"); } } } }
1 回答
蝴蝶不菲
TA貢獻1810條經(jīng)驗 獲得超4個贊
您需要處理可能是不同大小寫等的用戶輸入。因此，最簡單的方法是只訪問隨機單詞中的每個字符一次。
這是我為解決此問題而制作的 REPL：
using System;
using System.Collections.Generic;
class MainClass {
public static void Main (string[] args) {
var word = "Tomato";
var input = "t";
var letter = input.ToLower()[0];
var indices = new List<int>();
for(var i = 0; i < word.Length; i++)
if (word.ToLower()[i] == letter)
indices.Add(i);
Console.WriteLine($"Secret word: {word}");
Console.WriteLine($"User guess: {input}");
Console.WriteLine($"Found at {String.Join(", ", indices)}");
}
}
及其輸出：
Mono C# compiler version 4.0.4.0
Secret word: Tomato
User guess: t
Found at 0, 4
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