Root base class in C++

2020-03-01 15:57发布

Every object in .NET inherits (directly or indirectly) from the common root base "Object". Is there such a common object root in C++? How do I pass any object to a function?

public void DoSomeStuff(object o) { ... }

EDIT: To clarify, the purpose: In that function I want to invoke a pointer to member function. For that I need the object instance and pointer to the required function. To simplify readability I want to make a wrapper containing the both required information. I'm not sure if that is the best way, but it is the background idea.

6条回答
兄弟一词,经得起流年.
2楼-- · 2020-03-01 16:22

There is no common base class; but using a something like boost::any or more generally a template based approach is preferred over a void*.

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等我变得足够好
3楼-- · 2020-03-01 16:24

There is no common base class in C++. However, there are already libraries that allow you to call member functions as delegates. You can try using boost::function together with boost::bind or boost::lambda.

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▲ chillily
4楼-- · 2020-03-01 16:30

For that I need the object instance and pointer to the required function.

That sounds a lot like "delegates". First, you definitely will need to define a common base class for all the objects that you want to call. In C++ you can use multiple inheritance if the object already belong to some other hierarchy.

Then have a read through Member Functions and the Fastest Possible C++ Delegates which is an excellent in-depth article on the topic of delegates (which are an object and member function pointer bound together). Using the header file described in that article, you can create delegates and easily call them just like regular function pointers.

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等我变得足够好
5楼-- · 2020-03-01 16:37

There is no common root class. Use either void* to pass any object into a function, or better define some base class.

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Rolldiameter
6楼-- · 2020-03-01 16:37

Template functions are present and avoid the need for such root parent of all classes.

template <class T>
void DoSomeStuff(T const &t) {
    // Do the stuff with t...
    t.callTheFunction();
}

If all your objects have a member function callTheFunction(), then you got the exactly same behavior than having a root base class, with the same requirement (all your classes have a function with that name).

In addition, you got the additional benefit of being able to specialize the template function DoSomeStuff() for some classes that are not yours, and could not inherit your virtual member function.

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爷、活的狠高调
7楼-- · 2020-03-01 16:43

Sorry - there is no root base object in C++. But you can define your own for all your classes.

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