如何在返回值爲該類的對象類型的類中創建成員函數？

Question

下面是問題：

定義一個名爲Month的類，它是一個月的抽象數據類型。你的班級 有一個int類型的成員變量來代表一個月份（一月份爲1，二月份爲 ，等等）。包含以下所有成員函數：使用構造函數 將月份名稱中的前三個字母設置爲三個 參數，使用整數作爲參數設置月份的構造函數（1代表 January，2代表二月等等），一個默認的構造函數，一個輸入函數，該函數將月份讀取爲一個整數，一個輸入函數將該月份作爲第一個月份的三個字母讀取月份，一個輸出函數輸出該月份 作爲整數，輸出函數將月份作爲月份的前三個字母 輸出，而將下一個月的月份返回爲 的成員函數。將你的類定義嵌入到測試程序中。

這裏是我的代碼：

#include <iostream> 
    #include <cstdlib> 
    #include <string> 
    using namespace std; 

    class Month 
    { 
    public: 
Month(char firstLetter, char secondLetter, char thirdLetter); 
//Initializes month according to arguments for first 3 letters 

Month(int month_number); 
//Initializes month according to argument for integer between 1 and 12 

Month(); 
//Initializes month to January 

void input_integer();//Reads month as integer 
void input_letter();//Reads month as first 3 letters in the name of month 
int letters_to_integer(char firstLetter, char secondLetter, char thirdLetter); 
//Converts first 3 letters of the month to integer equivalent of the month 
string integer_to_letters(int month_number); 
//Converts month integer to the first 3 letters of the name of the month 
void output_integer();//Outputs the month as an integer 
void output_letter();//Outputs the month as first 3 letters of the name of the month 

void next_month();//Returns the next month in letters 
    private: 
int month_number;//Represents the month as an integer 
string month_letters;//Represents the month in terms of its first 3 letters 
char firstLetter, secondLetter, thirdLetter; 
int next_month_number;//Represents the next month as an integer 
    }; 

    int main() 
    { 
Month month1('m', 'a', 'y'), month2(7), month3; 
cout<<"Object month1 is initialized as follows:\n"; 
month1.output_letter(); 
cout<<"Object month2 is initialized as follows:\n"; 
month2.output_letter(); 
cout<<"Object month3 is initialized as follows:\n"; 
month3.output_letter(); 

cout<<"Enter the first 3 letters of the month\n"; 
month3.input_letter(); 
cout<<"That month expressed as in integer is:\n"; 
month3.output_integer(); 
cout<<endl; 


cout<<"Enter the month as an integer between 1 and 12:\n"; 
month3.input_integer(); 
cout<<"That is equivalent to the month of:\n"; 
month3.output_letter(); 
cout<<"in letters\n"; 
cout<<"The next month is: \n"; 
month3.next_month(); 
cout<<endl; 

system("pause"); 
return 0; 
    } 

    Month::Month(char firstLetter, char secondLetter, char thirdLetter) 
    { 
month_number = letters_to_integer(firstLetter, secondLetter, thirdLetter); 
month_letters = integer_to_letters(month_number); 
    } 

    Month::Month(int month_number) 
    { 
month_letters = integer_to_letters(month_number); 
    } 

    Month::Month() 
    { 
month_number = 1; 
month_letters = "Jan\n"; 
    } 

    void Month::input_integer() 
    { 
cin>>month_number; 
month_letters = integer_to_letters(month_number); 
next_month_number = month_number + 1; 
    } 

    void Month::input_letter() 
    { 
cin>>firstLetter>>secondLetter>>thirdLetter; 
month_number = letters_to_integer(firstLetter, secondLetter, thirdLetter); 
    } 

    void Month::output_letter() 
    { 
cout<<month_letters; 
    } 

    void Month::output_integer() 
    { 
cout<<month_number; 
    } 

    int Month::letters_to_integer(char firstLetter, char secondLetter, char  thirdLetter) 
    { 
if ((firstLetter == 'j') && (secondLetter == 'a') && (thirdLetter == 'n')) 
{ 
    month_number = 1; 
} 
else if ((firstLetter == 'f') && (secondLetter == 'e') && (thirdLetter == 'b')) 
{ 
    month_number = 2; 
} 
else if ((firstLetter == 'm') && (secondLetter == 'a') && (thirdLetter == 'r')) 
{ 
    month_number = 3; 
} 
else if ((firstLetter == 'a') && (secondLetter == 'p') && (thirdLetter == 'r')) 
{ 
    month_number = 4; 
} 
else if ((firstLetter == 'm') && (secondLetter == 'a') && (thirdLetter == 'y')) 
{ 
    month_number = 5; 
} 
else if ((firstLetter == 'j') && (secondLetter == 'u') && (thirdLetter == 'n')) 
{ 
    month_number = 6; 
} 
else if ((firstLetter == 'j') && (secondLetter == 'u') && (thirdLetter == 'l')) 
{ 
    month_number = 7; 
} 
else if ((firstLetter == 'a') && (secondLetter == 'u') && (thirdLetter == 'g')) 
{ 
    month_number = 8; 
} 
else if ((firstLetter == 's') && (secondLetter == 'e') && (thirdLetter == 'p')) 
{ 
    month_number = 9; 
} 
else if ((firstLetter == 'o') && (secondLetter == 'c') && (thirdLetter == 't')) 
{ 
    month_number = 10; 
} 
else if ((firstLetter == 'n') && (secondLetter == 'o') && (thirdLetter == 'v')) 
{ 
    month_number = 11; 
} 
else if ((firstLetter == 'd') && (secondLetter == 'e') && (thirdLetter == 'c')) 
{ 
    month_number = 12; 
} 
return month_number; 
    } 

    string Month::integer_to_letters(int month_number) 
    { 
string month_letters; 

switch (month_number) 
{ 
    case 1: 
     month_letters = "Jan\n"; 
     break; 
    case 2: 
     month_letters = "Feb\n"; 
     break; 
    case 3: 
     month_letters = "Mar\n"; 
     break; 
    case 4: 
     month_letters = "Apr\n"; 
     break; 
    case 5: 
     month_letters = "May\n"; 
     break; 
    case 6: 
     month_letters = "Jun\n"; 
     break; 
    case 7: 
     month_letters = "Jul\n"; 
     break; 
    case 8: 
     month_letters = "Aug\n"; 
     break; 
    case 9: 
     month_letters = "Sep\n"; 
     break; 
    case 10: 
     month_letters = "Oct\n"; 
     break; 
    case 11: 
     month_letters = "Nov\n"; 
     break; 
    case 12: 
     month_letters = "Dec\n"; 
     break; 
    default: 
     month_letters = "Invalid integer for month"; 
     break; 

} 
return month_letters; 
    } 

    void Month::next_month() 
    { 
cout<<integer_to_letters(next_month_number); 
    } 

對不起它的縮進一切奇怪，但是這是我的。我的代碼運行得很好。我唯一不知道該怎麼做的是問題的最後一部分，它要求創建「一個會員函數，該函數返回下一個月的月份爲 的值。」我根本不理解這一點。所以在代碼中我只輸出下個月的字符串。請幫助我這部分。在課堂上，我們只講課，這是作業。請儘可能簡單和基本。我一般不太熟悉C++或編程。

謝謝。

Answer 1

「並返回下個月類型月份的值的成員函數」

所以基本上你的類Month需要有一個返回下一個月的Month類型的對象的功能。例如，如果我有一個Month等於「June」的對象，並且我調用了一個函數...將其稱爲getNextMonth()它應該返回一個Month對象等於「July」。

所以，你每月類需要有一個功能，想必定義爲Month getNextMonth();的public節中，我將離開你弄清楚如何定義的功能，但我會建議「複製構造函數」會發生什麼讀了當你通過值返回一個類實例時。你的類不需要任何變量的「深層拷貝」，所以默認的拷貝構造函數應該沒問題，但是當你的代碼運行時知道實際發生了什麼是很好的。

Answer 2

儘可能基本： 返回類型： - 當您定義或聲明函數ex-int main（）;時，在函數名稱前寫入。而不是使用void或int你的問題要求你使用返回類型作爲你的類，即你已經定義的'月'，所以不要返回任何東西或int，你將返回一個類的對象。Month