可以通过引用以基类作为参数的函数来传递派生类 [英] Is it possible to pass derived classes by reference to a function taking base class as a parameter
问题描述
假设我们有一个抽象基类 IBase
和纯虚拟方法(一个接口)。
Say we have an abstract base class IBase
with pure virtual methods (an interface).
从基类中派生 CFoo
, CFoo2
。
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,通常的情况是这样的:
The usual scenario in these cases is like this:
IBase *ptr = static_cast<IBase*>(new CFoo("abc"));
Foo(ptr);
delete ptr;
但是指针管理最好避免,所以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
你不必上传你的对象。
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.
所有:
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
当您必须下载时,您可能需要考虑使用 dynamic_cast
,如果你的类型是多态的。
When you have to downcast you may want to consider using dynamic_cast
, if your types are polymorphic.
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