从抽象基类的多个部分实现继承？

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可以有一些抽象接口的部分实现，然后收集这些部分实现到单个具体类通过使用多重继承？

Is it possible to have a number of partial implementations of an abstract interface, and then collect these partial implementations into a single concrete class by using multiple inheritence?

我有以下示例代码：

#include <iostream> struct Base { virtual void F1() = 0; virtual void F2() = 0; }; struct D1 : Base { void F1() override { std::cout << __func__ << std::endl; } }; struct D2 : Base { void F2() override { std::cout << __func__ << std::endl; } }; // collection of the two partial implementations to form the concrete implementation struct Deriv : D1, D2 { using D1::F1; // I added these using clauses when it first didn't compile - they don't help using D2::F2; }; int main() { Deriv d; return 0; }

无法编译时出现以下错误：

This fails to compile with the following errors:

main.cpp: In function ‘int main()’: main.cpp:27:11: error: cannot declare variable ‘d’ to be of abstract type ‘Deriv’ main.cpp:19:8: note: because the following virtual functions are pure within ‘Deriv’: main.cpp:5:18: note: virtual void Base::F1() main.cpp:6:18: note: virtual void Base::F2()

推荐答案

尝试从 Base ：

struct D1 : virtual Base { void F1() override { std::cout << __func__ << std::endl; } }; struct D2 : virtual Base { void F2() override { std::cout << __func__ << std::endl; } };

没有虚拟继承，你的多继承场景看起来像是继承自两个单独的和不完整的基类 D1 和 D2 ，两者都不能实例化。

Without the virtual inheritance, your multiple inheritance scenario looks like inheritance from two separate and incomplete base classes D1 and D2, neither of which can be instantiated.