從閱讀文件中對不同類型的多個數組進行排序

Question

我正在研究一個模仿「動物採用代理」的程序。我從文件中讀取，其中包含動物名稱，品種，年齡，價格和性別列表。我有兩個文件，一個是貓和狗。

用戶可以通過上面列出的類別對列表進行排序。我目前有一個for循環，可以精確地對他們選擇的類別進行排序;但是，其他類別不會相應地訂購。我不知道如何去做這件事。 下面是我的代碼的簡明版本，它只允許訪問狗部分並按名稱排序，而不是選擇如何排序。

#include <iostream> 
    #include <fstream> 
    #include <istream> 
    #include <cctype> 
    #include <string> 
    #include <string.h> 
    #include <cstring> 
    #include <iomanip> 
    #include <vector> 
    #include <algorithm> 
    using namespace std; 

    int animalMenu, animalCount, animAge[50], animPrice[50], entry = 0, total; 
    string animType, animName[50], animBreed[50], animSex[50], takeHomeWith; 
    ifstream animalInform; 
    const int WIDTH = 8, BIG_WIDTH = 12; 

    void sortingHat(string[]); 
    void innerSorting(string[], int); 

    int main() { 
      animalInform.open("Dog Information.txt"); 
      animType = "dogs"; 

      // SET NUMBER OF ANIMALS IN FILE 
      animalInform >> animalCount; 
      cout << "There are " << animalCount << " " << animType << "! \n"; 

      // SETS ALL THE VALUES BY READING FROM FILE 
      for (int entry = 0; entry < animalCount; entry++) { 
       animalInform >> animName[entry] >> animBreed[entry] >> animAge[entry] >> animPrice[entry] >> animSex[entry]; 
       cout << setw(BIG_WIDTH) << animName[entry] << setw(BIG_WIDTH) << animBreed[entry] << setw(WIDTH) << animAge[entry] << setw(WIDTH) << animPrice[entry] << setw(WIDTH) << animSex[entry] << endl; 
      } 
      // CLOSE FILE 
      animalInform.close(); 

      // CALL FUNCTION TO SORT (BY NAME ONLY) 
      sortingHat(animName); 
      cout << endl; 
      // DISPLAY NEWLY SORTED LIST 
      for (int entry = 0; entry < animalCount; entry++) { 
       cout << setw(BIG_WIDTH) << animName[entry] << setw(BIG_WIDTH) << animBreed[entry] << setw(WIDTH) << animAge[entry] << setw(WIDTH) << animPrice[entry] << setw(WIDTH) << animSex[entry] << endl; 
      } 
      system("pause"); 
    } 

    void sortingHat(string sortingString[]) 
    { // SORTS DATA AND PUTS IT IN ORDER, ALPHABETICAL -- 
     for (int outer = 0; outer <= animalCount; outer++) 
     { 
      for (int entry = 0; entry <= (animalCount - 2); entry++) { 
       string temporary[50]; 
       if (sortingString[entry] > sortingString[entry + 1]) 
        innerSorting(sortingString, entry); 
      } 
     } 
    } 

    void innerSorting(string sorter[], int entry) 
    { 
     string temporary[50]; 
     temporary[entry] = sorter[entry]; 
     sorter[entry] = sorter[entry + 1]; 
     sorter[entry + 1] = temporary[entry]; 
    } 

所以我顯然沒有任何東西可以讓其他條目跟隨。 所以，如果我選擇的名稱進行排序，我的輸出（這是寫在我的文件）會從

Brienne Shepard 6 $150 F 
Jon Labrador 3 $200 M 
Aemon ShihTzu 10 $50 M 

到

Aemon Shepard 6 $150 F 
Brienne Labrador 3 $200 M 
Jon ShihTzu 10 $50 M 

而且我希望它這樣做（如果選擇按名稱排序）：

Aemon ShihTzu 10 $50 M 
Brienne Shepard 6 $150 F 
Jon Labrador 3 $200 M 

Answer 1

如果我正確理解了你，你有一組包含動物特徵的數組。而且你將按照一種特性來排序所有數組。如果是這樣，那麼你可以爲所有數組寫一個通用函數。

例如

enum SortType { SortByName, /* other types of sorting */, SortByAge }; 

//... 

void bubble_sort(std::string animName[], 
        /* other characteristics */ 
        int animAge[], 
        size_t n, 
        SortType type) 
{ 
    for (size_t last; not (n < 2); n = last) 
    { 
     for (size_t i = last = 1; i < n; i++) 
     { 
      bool less = false; 
      switch (type) 
      { 
      case SortByName: 
       less = animName[i] < animName[i-1]; 
       break; 
      /* other cases */ 
      case SortByAge: 
       less = animAge[i] < animAge[i-1]; 
       break; 
      } 

      if (less) 
      { 
       /* swapping elements of all the arrays */ 
       last = i; 
      } 
     } 
    } 
} 

考慮到這個交換函數

void innerSorting(string sorter[], int entry) 
{ 
    string temporary[50]; 
    ^^^^^^^^^^^^^^^^^^^^ 
    temporary[entry] = sorter[entry]; 
    sorter[entry] = sorter[entry + 1]; 
    sorter[entry + 1] = temporary[entry]; 
} 

不應該使用一個字符串數組。它可以寫成像

void innerSorting(string sorter[], size_t entry) 
{ 
    string temporary; 
    ^^^^^^^^^^^^^^^^^ 
    temporary = sorter[entry]; 
    sorter[entry] = sorter[entry + 1]; 
    sorter[entry + 1] = temporary; 
}