Question

我有2个课程，A和B。在A我有3个私人字段。在类B中，我想编写一个复制构造函数，并初始化类A中的私有字段。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;
}

我明白当我从private更改为protected时，它应该可以工作（当然也可以工作） - 但是如何处理我的代码中的情况？问题是：如何在复制构造函数中初始化基类中的私有字段？我使用当前代码得到以下错误：

/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

您需要做的就是在复制构造A时调用B的复制构造函数，如

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

由于B继承自A，因此这是合法的，A将复制A的{{1}}部分。

另请注意，所有这些指针都是不必要的，并使代码更复杂。我们可以改写它：

other

Answer 2

您需要从复制构造函数中调用父复制构造函数，为其提供相同的参数。

继承和复制构造函数 - 如何从基类初始化私有字段？

2 个答案: