C++ 使用公共函数签名实现多个接口的派生类
当我试图编译我的代码时,我得到了一个编译错误。 错误是:C++ 使用公共函数签名实现多个接口的派生类,c++,class,inheritance,multiple-inheritance,C++,Class,Inheritance,Multiple Inheritance,当我试图编译我的代码时,我得到了一个编译错误。 错误是: multi.cc: In function ‘int main()’: multi.cc:35: error: cannot declare variable ‘mdc’ to be of abstract type ‘MostDerivedClass’ multi.cc:27: note: because the following virtual functions are pure within ‘MostDerivedClas
multi.cc: In function ‘int main()’:
multi.cc:35: error: cannot declare variable ‘mdc’ to be of abstract type ‘MostDerivedClass’
multi.cc:27: note: because the following virtual functions are pure within ‘MostDerivedClass’:
multi.cc:13: note: virtual int Interface2::common_func()
multi.cc:36: error: request for member ‘common_func’ is ambiguous
multi.cc:13: error: candidates are: virtual int Interface2::common_func()
multi.cc:21: error: virtual int InterimClass::common_func()
这是我的代码:
class Interface1 {
public:
virtual int common_func() = 0;
virtual ~Interface1() {};
};
class Interface2 {
public:
virtual int common_func() = 0;
virtual int new_func() = 0;
virtual ~Interface2() {};
};
class InterimClass : public Interface1 {
public:
virtual int common_func() {
return 10;
}
};
class MostDerivedClass : public InterimClass, public Interface2 {
public:
virtual int new_func() {
return 20;
}
};
int main() {
MostDerivedClass mdc;
int x = mdc.common_func();
cout << "The value = " << x << endl;
Interface2 &subset_of_funcs = dynamic_cast<Interface2 &>(mdc);
x = subset_of_funcs.common_func();
}
类接口1{
公众:
虚int公共函数()=0;
虚拟~Interface1(){};
};
类接口2{
公众:
虚int公共函数()=0;
虚拟int new_func()=0;
虚拟~Interface2(){};
};
类间类:公共接口1{
公众:
虚int公共函数(){
返回10;
}
};
类MostDerivedClass:公共InterimClass,公共接口2{
公众:
虚拟int new_func(){
返回20;
}
};
int main(){
最高级mdc;
int x=mdc.common_func();
您是否需要在MostDerivedClass
中定义一个common_func()
来满足您从Interface2
继承的需求
你可以试试类似的东西
virtual int common_func() {
return InterimClass::common_func();
}
如果无法更改第一个接口1
如果你想在类之间建立真正的继承关系,你需要遵循Lol4t0建议。从Interface1
提取一个超类,并为这个新创建的类创建Interface2
子类。示例:
class RootInterface{
public :
virtual int common_func() = 0;
virtual ~RootInterface(){}
};
class Interface1 : public virtual RootInterface{
public:
virtual ~Interface1() {};
};
class Interface2 : public virtual RootInterface{
public:
virtual int new_func() = 0;
virtual ~Interface2() {};
};
class InterimClass : public Interface1 {
public:
virtual int common_func() {
return 10;
}
};
class MostDerivedClass : public InterimClass, public Interface2 {
public:
virtual int new_func() {
return 20;
}
};
在MostDerivedClass中添加一个重写,并从中调用InterimClass::common_func()。首先,我不太理解您的代码的含义
您需要知道,只实现了Interface1::common_func
为什么不让Interface2从Interface1继承?我想您希望两个常用的函数方法相等
示例代码(使用多态性):
类接口1
{
公众:
虚int公共函数()=0;
虚拟~Interface1(){};
};
类接口2:公共接口1{
公众:
虚int公共函数()=0;
虚拟int new_func()=0;
虚拟~Interface2(){};
};
类间类:公共接口2{
公众:
虚int公共函数(){
返回10;
}
};
类MostDerivedClass:公共InterimClass{
公众:
虚拟int new_func(){
返回20;
}
};
int test_func()
{
接口1*i1=新的MostDerivedClass;
int x=i1->common_func();
不能从第一个接口派生第二个接口,从第二个接口中删除virtual int common_func()=0;
的声明,并使用关键字virtual引导编译器实现
class Interface1 {
public:
virtual int common_func() = 0;
virtual ~Interface1() {};
};
class BaseClass : public virtual Interface1 {
public:
virtual int common_func() {
return 10;
}
};
class Interface2 : public virtual Interface1{
public:
virtual int new_func() = 0;
virtual ~Interface2() {};
};
class DerivedClass : public virtual BaseClass, public virtual Interface2 {
public:
virtual int new_func() {
return 20;
}
};
您的公用函数实际上不是“公用的”。Interface1::common_func()
与Interface2::common_func()没有任何共同之处
。如果您想要真正的通用函数,您应该从Interface1
派生Interface2
。然后应用@juanchopanza答案。感谢您的回答。我只是对现有代码的更改犹豫不决。希望以后某个时候,我可以重构代码以反映新的继承结构。
class Interface1 {
public:
virtual int common_func() = 0;
virtual ~Interface1() {};
};
class BaseClass : public virtual Interface1 {
public:
virtual int common_func() {
return 10;
}
};
class Interface2 : public virtual Interface1{
public:
virtual int new_func() = 0;
virtual ~Interface2() {};
};
class DerivedClass : public virtual BaseClass, public virtual Interface2 {
public:
virtual int new_func() {
return 20;
}
};