Question

I'm having trouble where a user cannot use the same word after input. Below is the snippet of a Boggle program I'm working on.

The code below does not work the way I want it to if for example:

(1)The user inputs the word: fun --> store into temp[0] --> increment count

(2)The user inputs the word: fun --> repeat while loop input for another different input

(3)The user inputs the word: bye --> break while loop --> store into temp[1] --> increment count

(4)The user inputs the word: bye or fun --> will repeat for another input

(5)The user inputs the word: good --> store into temp[2] --> increment count

(6)The user inputs the word: bye --> repeat while loop for another input,

(7)The user inputs the word: fun --> While loop breaks and the word becomes VALID.

You see.. the problem is that it doesn't loop through back to temp[0] to find the word fun again to say its invalid.

Any assistance with this would be appreciated.

//seconds is used with time library, but just ignore its declaration
while (seconds < 300){

   string word;
   string temp[100];
   int count = 0;


   cout << "Enter:" << endl;
   cin >> word;

   for (int i = 0; i < count; i++){
      while (word == temp[i]){
        cout << "Word was already used. Please type another word." << endl;
        cin >> word;
      }
   }

   temp[count] = word;
   count = count + 1;

   if (seconds >= 300)
      break;
}

Answer 1

试试这个：

string word;
string temp[100];
int count = 0;

cout << "Enter:" << endl;
bool found;
do
{
    cin >> word;

    found = false;
    for (int i = 0; i < count; i++)
    {
        if (temp[i] == word)
        {
            found = true;
            cout << "Word was already used. Please type another word." << endl;
            break;
        }
    }
}
while (found);

temp[count++] = word;

如果在word中的任何位置找到temp，它将会询问一个新单词，然后每次都重新检查整个temp数组。您的原始代码没有这样做，并且只检查当前索引i，即使在请求新输入时也是如此。

Using the inputed word only once

1 个答案: