我正在学习C ++,并且对多重继承和接口类感到困惑。
我想拥有一个从其他几个继承的类。另外,我想通过接口使用该派生类。所以我想,派生类应该扩展基类,派生接口应该扩展基接口。我本来可以用其他语言完成的,但是我认为C ++不能那样工作。
这是我认为应该工作的代码:
#include <iostream>
using std::cout;
using std::endl;
class
Base1Itf
{
public:
virtual void blue() = 0;
};
class
Base1Abs
:
public Base1Itf
{
public:
void blue()
{
cout << "blue" << endl;
}
};
class
DerivedItf
:
public Base1Itf
{
public:
virtual void red() = 0;
};
class
Derived
:
public Base1Abs,
public DerivedItf
{
public:
void red()
{
cout << "red" << endl;
}
};
int main()
{
DerivedItf* d = new Derived();
d->red();
d->blue();
delete d;
return 0;
}
这是我得到的编译器错误:
src/test.cpp: In function ‘int main()’:
src/test.cpp:49:30: error: invalid new-expression of abstract class type ‘Derived’
DerivedItf* d = new Derived();
^
src/test.cpp:35:2: note: because the following virtual functions are pure within ‘Derived’:
Derived
^~~~~~~
src/test.cpp:10:16: note: virtual void Base1Itf::blue()
virtual void blue() = 0;
^~~~
在该示例中,仅实现了一个基类,但还会有更多的基类。
我做错了什么?谢谢。
编辑
如果为了避免钻石问题而删除了Base1Itf的Base1Abs继承,则编译器会显示相同的错误。
这是C ++中众所周知的钻石问题。这是您解决的方法:
#include <iostream>
using std::cout;
using std::endl;
class Base1Itf {
public:
virtual void blue() = 0;
};
class Base1Abs : virtual public Base1Itf {
public:
void blue() {
cout << "blue" << endl;
}
};
class DerivedItf : virtual public Base1Itf
{
public:
virtual void red() = 0;
};
class Derived : public Base1Abs, public DerivedItf
{
public:
void red() {
cout << "red" << endl;
}
};
int main()
{
Derived* d = new Derived();
d->red();
d->blue();
delete d;
return 0;
}