c – 如何覆盖另一个基类中的函数?
作者:互联网
我不确定要使用的术语,但这是我的例子:
class Base {
public:
virtual void test() = 0;
};
class Mixin {
public:
virtual void test() { }
};
class Example : public Base, public Mixin {
};
int main(int argc, char** argv) {
Example example;
example.test();
return 0;
}
我希望我的Mixin类实现纯虚函数Base :: test,但是当我编译它时,它说:
test.cpp: In function ‘int main(int, char**)’:
test.cpp:15:13: error: cannot declare variable ‘example’ to be of abstract type ‘Example’
Example example;
^
test.cpp:11:7: note: because the following virtual functions are pure within ‘Example’:
class Example : public Base, public Mixin {
^
test.cpp:3:18: note: virtual void Base::test()
virtual void test() = 0;
^
test.cpp:16:13: error: request for member ‘test’ is ambiguous
example.test();
^
test.cpp:8:18: note: candidates are: virtual void Mixin::test()
virtual void test() { }
^
test.cpp:3:18: note: virtual void Base::test()
virtual void test() = 0;
^
我可以添加一个using语句,使其不含糊:
class Example : public Base, public Mixin {
public:
using Mixin::test;
};
但它说我还没有实现它:
test.cpp: In function ‘int main(int, char**)’:
test.cpp:17:13: error: cannot declare variable ‘example’ to be of abstract type ‘Example’
Example example;
^
test.cpp:11:7: note: because the following virtual functions are pure within ‘Example’:
class Example : public Base, public Mixin {
^
test.cpp:3:18: note: virtual void Base::test()
virtual void test() = 0;
^
是否有可能做到这一点?
我知道一个选项是让Mixin从Base继承,但在我的情况下,有几个派生类,它们不共享一个共同的祖先.
解决方法:
您不能直接让类覆盖不是其基类的方法.但是你可以用迂回的方式来做.我将介绍两种这样的方法 – 我更喜欢第二种方法.
方法1
Daniel Paul在thinkbottomup.com.au的一篇文章中描述了这一点,题为C++ Mixins – Reuse through inheritance is good… when done the right way.
在你的情况下,这是它的样子:
class Base {
public:
virtual void test() = 0;
};
template <typename T>
class Mixin : public T {
public:
virtual void test() override { /*... do stuff ... */ }
};
class UnmixedExample : public Base {
/* definitions specific to the Example class _not_including_
a definition of the test() method */
};
using Example = class Mixin<UnmixedExample>;
int main(int argc, char** argv) {
Example{}.test();
return 0;
}
方法2:CRTP!
CRTP是“奇怪的重复模板模式” – 如果您之前没有看过它,请务必遵循该链接.使用这种方法,我们将使用虚拟继承说明符来避免歧义,与前一种方法不同 – 我们不会颠倒Mixin和Example类的继承顺序.
class Base {
public:
virtual void test() = 0;
};
template <typename T>
class Mixin : virtual T {
public:
virtual void test() override { /*... do stuff ... */ }
};
class Example : public virtual Base, public virtual Mixin<Base> {
/* definitions specific to the Example class _not_including_
a definition of the test() method */
};
int main(int argc, char** argv) {
Example{}.test();
return 0;
}
注意两种解决方案:
>好奇CRTP是如何在整个地方重复出现的?
标签:c,inheritance,multiple-inheritance 来源: https://codeday.me/bug/20191008/1870840.html