2016-01-01 84 views
0

我正在寫一個代碼,對於一行或多行字符串,查找整體輸入是否只有「酷」(它的第一個中間和最後一個字符串是相同的)行,只有「 uncool「系列或兩者的組合。在特定情況下跳出外部循環

我遇到的問題是每當我輸入偶數循環終止。調試我發現,在跳出之前n得到值0,但我不明白這將如何使循環結束。

這是代碼:

#include <iostream> 
using namespace std; 

int main() { 
    // Bool has control if we have found a cool line/non-cool line 
    bool cool = false; 
    bool uncool = false; 

    int n; //lenght of input 
    while (cin >> n) { 
     if (cool and uncool) break; // we have found one of each so we know it is a mixed input 
     else if (n%2 == 0) uncool = true; // if the lenght is even there is no middle string 
     else { 
      // we are trying to see if the middle and last string are equal to the first 
      string comparing_string; 
      cin >> comparing_string; 

      string rest_of_sequence; 
      bool this_is_cool = true; 
      for (int i = n-2; i >= 0; i--) { // we input the rest of strings and compare them to the first 
       cin >> rest_of_sequence; 
       if ((i == n/2 or i == 0) and rest_of_sequence != comparing_string) this_is_cool = false; 
      } 

      if (this_is_cool) cool = true; 
      else uncool = true; 
     } 
    } 

    if (cool and uncool) cout << "both types" << endl; 
    else if (cool and not uncool) cout << "all cool" << endl; 
    else if (uncool and not cool) cout << "none cool" << endl; 
} 

任何幫助表示讚賞!我目前在uni的第一年,並始終打開推薦的書籍/網頁/視頻,以繼續學習:)

+0

替換if(cool and uncool)break; to if(cool && uncool)break; 這可能會幫助 –

+2

@Saiful爲什麼會這樣? – LogicStuff

+2

@SaifulIslam:並非如此,因爲['和'意味着'&&'](https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#C.2B.2B_operator_synonyms)。 – Cornstalks

回答

2

問題是我認爲程序會忽略在while循環中不是整數的輸入,但事實並非如此。

現在的代碼是正確的:

else if (n%2 == 0) {// if the lenght is even there is no middle string 
      uncool = true; 
      string just_passing_input; 
      for (int i = n; i > 0; i--) cin >> just_passing_input; 
     } 

感謝您的幫助的反饋,現在我要繼續學習。