繼承和複製構造函數 - 如何從基類初始化私有字段？

Question

我有2個班級，A和B。在A我有3個私人領域。在類B中，我想寫一個拷貝構造函數，並初始化類A的私有字段。然而，這不起作用：

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

class A 
{ 
    private: 

     string *field1; 
     string *field2; 
     string *field3; 
     double num1; 

    public: 

     A(string *o, string *n, string *m, double a=0) 
     { 
      field1 = new string(*o); 
      field2 = new string(*n); 
      field3 = new string(*m); 
      num1 = a; 
     } 

     A(const A& other) { 
      field1 = new string(*other.field1); 
      field2 = new string(*other.field2); 
      field3 = new string(*other.field3); 
      num1 = other.num1; 
     } 

     void show() 
     { 
      cout << *field1 << " " << *field2 << " " << *field3 << "\n"; 
     } 

     ~A() 
     { 
      delete field1; 
      delete field2; 
      delete field3; 
     } 
}; 

/*--------------------------------------------------------------------------------------------*/ 

class B : public A 
{ 
    private : 

     double num2; 
     double num3; 

    public: 

     B(double num2, double num3, string *o, string *n, string *num, double a=0) : A(o,n,num,a) 
     { 
      this->num2 = num2; 
      this->num3 = num3; 
     } 

     B(const B& other) : A(other.field1, other.field2, other.field3, other.num1) 
     { 
      num2 = other.num2; 
      num3 = other.num3; 
     } 

     void show() 
     { 
      cout << num2 << " " << num3 << "\n"; 
      A::show(); 
     } 
}; 

int main() 
{ 
    string o = "TEXT 111"; 
    string *optr = &o; 

    string n = "TEXT 222"; 
    string *nptr = &n; 

    string *numptr = new string("9845947598375923843"); 

    A ba1(optr, nptr, numptr, 1000); 
    ba1.show(); 

    A ba2(ba1); 
    ba2.show(); 

    A ba3 = ba2; 
    ba3.show(); 

    B vip1(20, 1000, optr, nptr, numptr, 3000); 
    vip1.show(); 

    B vip2(vip1); 
    vip2.show(); 

    delete numptr; 
    return 0; 
} 

我明白，當我從private到protected改變它應該工作（和作品，當然） - 但是如何處理，我在我的代碼的情況呢？問題是：如何在複製構造函數中初始化基類中的私有字段？我收到以下錯誤與當前代碼：

/home/yak/test.cpp|9|error: ‘std::string* A::field1’ is private| 
/home/yak/test.cpp|61|error: within this context| 
/home/yak/test.cpp|10|error: ‘std::string* A::field2’ is private| 
/home/yak/test.cpp|61|error: within this context| 
/home/yak/test.cpp|11|error: ‘std::string* A::field3’ is private| 
/home/yak/test.cpp|61|error: within this context| 
/home/yak/test.cpp|12|error: ‘double A::num1’ is private| 
/home/yak/test.cpp|61|error: within this context| 
||=== Build failed: 8 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===| 

Answer 1

所有你需要做的是調用A拷貝構造函數，當您複製構造B像

B(const B& other) : A(other) 
{ 
    num2 = other.num2; 
    num3 = other.num3; 
} 

由於B繼承A這是法律和A將複製的other的A部分。

還要注意，所有這些指針都是不必要的，使代碼更復雜。我們可以把它想：

class A 
{ 
private: 
    string field1; 
    string field2; 
    string field3; 
    double num1; 

public: 
    A(const string& o, const string& n, const string& m, double a = 0) : field1(o), field2(n), feild3(m), num1(a) {} 

    A(const A& other) field1(other.field1), field2(other.field2), feild3(other.feild3), num1(other.num1) {} 

    void show() 
    { 
     cout << field1 << " " << field2 << " " << field3 << "\n"; 
    } 
}; 

/*--------------------------------------------------------------------------------------------*/ 

class B : public A 
{ 
private: 

    double num2; 
    double num3; 

public: 

    B(double num2, double num3, const string& o, const string& n, const string& m, double a = 0) : A(o, n, num, a), num2(num2), num3(num3) {} 

    B(const B& other) : A(other), num2(other.num2), num3(other.num3) {} 

    void show() 
    { 
     cout << num2 << " " << num3 << "\n"; 
     A::show(); 
    } 
}; 

Answer 2

你需要從你的拷貝構造函數調用父拷貝構造函數，給它同樣的論點。

Answer 3

當使用B類的拷貝構造函數，你應該調用類A的拷貝構造函數

因此，通過這一個等效代碼：

B(const B& other) : A(other) 
    { 
     num2 = other.num2; 
     num3 = other.num3; 
    } 

另外，將類A的析構函數聲明爲虛函數。我想你也想爲show（）方法做到這一點。