c++ - 继承和复制构造函数——如何从基类初始化私有(private)字段？

Question

我有 2 个类，A 和 B。在 A 中，我有 3 个私有(private)字段。在 B 类中，我想编写一个复制构造函数，并从 A 类中初始化私有(private)字段。但是，这不起作用:

#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;
}

我确实明白，当我从 private 更改为 protected 时，它应该可以工作(当然可以工作)——但是如何处理我遇到的情况我的代码？问题是:如何在复制构造函数中初始化基类的私有(private)字段？当前代码出现以下错误:

/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

你需要做的就是在复制B时调用A的复制构造函数

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

因为 B 继承自 A 这是合法的并且 A 将复制 的 A 部分其他.

另请注意，所有这些指针都是不必要的，并且会使代码更加复杂。我们可以像这样重写它:

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();
    }
};