Question

在此程序中，我试图在此程序中灌输混合继承，但这给出了错误的输出。我已将算术作为基类，并将add，sub，mul，div作为其派生类。

然后我使用add，sub，mul，div作为基类派生了一个类结果。我已经尝试了所有数据类型，但是所有这些数据类型都会给出错误或为零的输出。

#include "iostream"
using namespace std;

class arithmetic
{
public:
    float var1,var2;
    void introduce()
    {
        cout<<"This program will perform arithmetic on two variables"<<endl
        <<"Enter the first variable: ";
        cin>>var1;
        cout<<"Enter the second variable: ";
        cin>>var2;
    }
};

class add:public arithmetic
{
protected:
    float res_add;
public:
    void show_add()
    {
        res_add=var1+var2;

        cout<<"Addition of those variables gives "<<res_add<<endl;
    }
};
class sub:public arithmetic
{
protected:
    float res_sub;
public:
    void show_sub()
    {
        res_sub=var1-var2;

        cout<<"Subtraction of those variables gives "<<res_sub<<endl;
    }
};
class mul:public arithmetic
{
protected:
    float res_mul;
public:
    void show_mul()
    {
        res_mul=var1*var2;

        cout<<"Multiplication of those variables gives "<<res_mul<<endl;
    }
};
class div:public arithmetic
{
protected:
    float res_div;
public:
    void show_div()
    {
        res_div=var1/var2;

        cout<<"Divison of those variables gives "<<res_div<<endl;
    }
};
class result:public add, public sub,public mul,public div
{
public:
    void showres()
    {
        cout<<"Arithmetic on the given two varibales gives us the following result:"<<endl;
    }
};

int main()
{
    result example;
    arithmetic var;
    var.introduce();
    example.showres();
    example.show_add();
    example.show_sub();
    example.show_mul();
    example.show_div();
    return 0;
}

Answer 1

您正在创建两个单独的对象var（基类对象）和example（派生类对象）。通过在示例对象中调用var.introduce（）并调用方法show_add（），show_sub（）等来初始化var1和var2，在示例对象var1和var2中未初始化。因此，您要调用的任何算术运算都将应用于示例对象的未初始化的var1和var2成员变量。

您不需要创建基类对象（var）。从示例中调用intruduce（）方法，它将开始正常工作。

请仔细阅读以下示例代码，以了解虚拟基类的概念。

#include <iostream>

class A 
{ 
    public: 
        int i; 
};

class B : virtual public A 
{ 
    public: 
        int j; 
};

class C: virtual public A 
{ 
    public: 
       int k; 
};

class D: public B, public C 
{ 
    public: 
        int sum; 
};

int main() 
{ 
    D ob; 
    ob.i = 10; //unambiguous since only one copy of i is inherited. 
    ob.j = 20; 
    ob.k = 30; 
    ob.sum = ob.i + ob.j + ob.k; 
    std::cout << "Value of i is : "<< ob.i<<"\n"; 
    std::cout << "Value of j is : "<< ob.j<<"\n";
    std::cout << "Value of k is : "<< ob.k<<"\n"; 
    std::cout << "Sum is : "<< ob.sum <<"\n"; 

    return 0; 
}

输出：

Value of i is : 10
Value of j is : 20
Value of k is : 30
Sum is : 60

Live Demo on coliru

Answer 2

如果要避免虚拟继承并希望简单一些，请使用组合。
为此，class result应该包含add，sub，mul和div的对象。

result的结果代码如下：

class result:public arithmetic
{
public:

    mul m;
    add a;
    sub s;
    div d;  
    void assignvals() 
    {
        m.var1 = var1; m.var2 = var2;
        a.var1 = var1; a.var2 = var2;
        s.var1 = var1; s.var2 = var2;
        d.var1 = var1; d.var2 = var2;
    }
    void showres()
    {               
        cout<<"Arithmetic on the given two variables gives us the following result:"<<endl;
    }
};

main的结果代码如下：

int main()
{
    result example;    
    example.introduce();
    example.assignvals();
    example.showres();
    example.a.show_add();
    example.s.show_sub();
    example.m.show_mul();
    example.d.show_div();
    return 0;
}

注意：：某些编译器会抱怨将div用作类名，因为它也是C ++中库函数的名称。因此，您将不得不更改此类的名称。

为什么此混合继承程序输出错误？

2 个答案: