一個類中的多個對象的構造函數

Question

所以我寫的程序需要我使用構造函數輸入對象（流動站）的初始值，我無法弄清楚正確的語法以便它讓我輸入每個輸入座標和方向的步驟。如果任何人能夠告訴我如何做到這一點，將不勝感激。

class Rover{ 

private: 

    string name; 
    int xpos; 
    int ypos; 
    string direction; //Use Cardinal Directions (N,S,E,W) 
    int speed; //(0-5 m/sec) 

public: 
    //Constructors 
    Rover(int,int,string,int); 
}; 
Rover::Rover(int one, int two, string three, int four) 
    { 
     cout<<"Please enter the starting X-position: "; 
     cin>>one; 
     cout<<"Please enter the starting Y-position: "; 
     cin>>two; 
     cout<<"Please enter the starting direction (N,S,E,W): "; 
     cin>>three; 
     cout<<"Please enter the starting speed (0-5): "; 
     cin>>four; 

     xpos=one; 
     ypos=two; 
     direction=three; 
     speed=four; 

     cout<<endl; 
    } 

int main(int argc, char** argv) { 

    int spd; 
    string direct; 
    string nme; 
    int x; 
    int y; 


    Rover r1(int x,int y, string direct, int spd); 
    Rover r2(int x,int y, string direct, int spd); 
    Rover r3(int x,int y, string direct, int spd); 
    Rover r4(int x,int y, string direct, int spd); 
    Rover r5(int x,int y, string direct, int spd); 






    return 0; 
} 

Answer 1

您的構造函數的實現應該比這更簡單。只需使用輸入參數初始化成員變量即可。

Rover::Rover(int xp, int yp, string dir, int sp) : 
    xpos(xp), ypos(yp), direction(dir), speed(sp) {} 

讀取輸入然後使用輸入構建對象的代碼可以移動到不同的函數，最好是非成員函數。

Rover readRover() 
{ 
    int xp; 
    int yp; 
    string dir; 
    int sp; 

    cout << "Please enter the starting X-position: "; 
    cin >> xp; 

    cout << "Please enter the starting Y-position: "; 
    cin >> xp; 

    cout << "Please enter the starting direction (N,S,E,W): "; 
    cin >> dir; 

    cout << "Please enter the starting speed (0-5): "; 
    cin >> sp; 

    // Construct an object with the user input data and return it. 
    return Rover(xy, yp, dir, sp); 
} 

，然後從main使用readRover多次如你所願。

Rover r1 = readRover(); 
Rover r2 = readRover(); 
Rover r3 = readRover(); 
Rover r4 = readRover(); 
Rover r5 = readRover(); 

函數readRover假定用戶提供了正確的輸入。如果用戶沒有提供正確的輸入，你的程序將被卡住，唯一的出路將是使用Ctrl + C或其他等價物來中止程序。

要處理用戶錯誤，您需要添加代碼來首先檢測錯誤，然後決定如何處理錯誤。

if (!(cin >> xp)) 
{ 
    // Error in reading the input. 
    // Decide what you want to do. 
    // Exiting prematurely is an option. 
    std::cerr << "Unable to read x position." << std::endl; 
    exit(EXIT_FAILURE); 
}