是否可以通过引用以基类为参数的函数来传递派生类
假设我们有一个带有纯虚方法(一个接口)的抽象基类 IBase.
Say we have an abstract base class IBase with pure virtual methods (an interface).
然后我们从基类派生CFoo、CFoo2.
Then we derive CFoo, CFoo2 from the base class.
我们有一个知道如何使用 IBase 的函数.
And we have a function that knows how to work with IBase.
Foo(IBase *input);
在这些情况下通常的场景是这样的:
The usual scenario in these cases is like this:
IBase *ptr = static_cast<IBase*>(new CFoo("abc"));
Foo(ptr);
delete ptr;
但是最好避免指针管理，那么在这种情况下有没有办法使用引用?
But pointer management is better to be avoided, so is there a way to use references in such scenario?
CFoo inst("abc");
Foo(inst);
其中 Foo 是:
Foo(IBase &input);
推荐答案
是的.您不必向上转换您的对象.必要时，所有对派生类型的引用/指针都会隐式转换为基对象引用/指针.
Yes. You don't have to upcast your objects. All references/pointers to derived types are converted implicitly to base objects references/pointers when necessary.
所以:
IBase* ptr = new CFoo("abc"); // good
CFoo* ptr2 = static_cast<CFoo*>(ptr); // good
CFoo* ptr3 = ptr; // compile error
CFoo instance("abc");
IBase& ref = instance; // good
CFoo& ref2 = static_cast<CFoo&>(ref); // good
CFoo& ref3 = ref; // compile error
当您不得不向下转型时，如果您的类型是多态的，您可能需要考虑使用 dynamic_cast.
When you have to downcast you may want to consider using dynamic_cast, if your types are polymorphic.
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