Hang子手遊戲 - 同時循環條件「包含在陣列中」

Question

我正在創建一個hang子手遊戲。我有一個問題，要求用戶重新輸入一個新的猜測（char），如果他們已經輸入了同一個（這些包含在字母[]數組中）。我已經開始了一個while循環，要求用戶重新輸入，如果「猜測」匹配存儲在數組中的一個字符。謝謝。

void play(string word) 
{ 
    string copy = word; 

    for(int i = 0; i < word.length(); i++) 
    { 
     copy[i] = ' '; 
    } 

    bool match = false; 
    bool valid = false; 
    int chance = 5; 
    char letters[26]; 
    int counter = 0; 
    char guess; 
    while (chance > 0 && match == false) 
    { 
     int blanks = 0; 
     for (int i = 0; i < word.length(); i++) 
     { 
      if(copy[i] == ' ') 
       blanks++; 
     } 

     cout << '|'; 
     for(int i = 0; i < word.length(); i++) 
     { 
      cout << copy[i] << '|'; 
     } 
     cout << "\t(There are '" << blanks << "' blanks)" << endl; 
     cout << "Incorrect letters guessed: "; 

     for (int i = 0; i<counter; i++) 
     { 
      cout << "'" << letters[i] << "' "; 
     } 

     cout << "\n\nEnter guess: "; 
     cin >> guess; 
     for(int i = 0; i < counter; i++) 
     { 
      if(guess == letters[i]) 
       valid = true; 
     } 

     while (valid == true) 
     { 

      for(int i = 0; i < counter; i++) 
      { 
       if(guess == letters[i]) 
        valid = true; 
       else 
        valid = false; 
      } 
      if (valid) 
      { 
       cout << "\n\nAlready tried that one. Enter guess: "; 
       cin >> guess; 
      } 

     } 
     long find = word.find(guess); 

     if (find != string::npos) 
     { 
      copy[find] = word[find]; 
      for (int i = 0; i < word.length(); i++) 
      { 
       if (word[i] == word[find]) 
       { 
        copy[i] = word[find]; 
       } 
      } 
     } 



     else 
     { 
      chance--; 

      letters[counter] = guess; 
      counter++; 
      printMan(chance); 
     } 



     if (copy == word) 
     { 
      match = true; 
     } 
    } 

    if (match == true) 
     cout << "\nYou saved a life! You managed to win! Congrats, cheater." << endl; 
    else 
     cout << "\nThe man has been hung! You LOSE! The word was '" << word << "', you idiot." << endl; 

} 

Answer 1

for(int i = 0; i < counter; i++) 
     { 
      if(guess == letters[i]) 
       valid = true; 
      else 
       valid = false; 
     } 

這個for循環的問題是，你找到一個有效的猜測之後......你不停止循環。

你繼續檢查字母是否等於循環中的下一個值 - 下一個和下一個。和valid不斷覆蓋，直到你到達循環結束 - 此時valid等於檢查值對最後信letters。

因此，如果guess與letters中的最後一個字母匹配，則只會調用您的「檢查猜測」代碼。

您需要查看在找到匹配猜測的時刻如何退出循環。