C ++：如何将一个派生类的容器传递给一个需要它们的基类的容器的函数？ [英] C++: How do I pass a container of derived classes to a function expecting a container of their base classes?

问题描述

HI！任何人都知道我可以在下面的代码中行 chug（derlist）; 工作？

HI! Anyone know how I can make the line "chug(derlist);" in the code below work?

#include <iostream>
#include <list>
using namespace std;

class Base
{
public:
    virtual void chug() { cout << "Base chug\n"; }
};

class Derived : public Base
{
public:
    virtual void chug() { cout << "Derived chug\n"; }
    void foo() { cout << "Derived foo\n"; }
};

void chug(list<Base*>& alist)
{
    for (list<Base*>::iterator i = alist.begin(), z = alist.end(); i != z; ++i)
        (*i)->chug();
}

int main() 
{
    list<Base*> baselist;
    list<Derived*> derlist;

    baselist.push_back(new Base);
    baselist.push_back(new Base);
    derlist.push_back(new Derived);
    derlist.push_back(new Derived);

    chug(baselist);
    // chug(derlist);  // How do I make this work?

    return 0;
}

我需要这个的原因基本上是，我有一个非常复杂的对象，我需要传递给某些函数，只关心这些复杂对象中的一个或两个虚拟函数。

The reason I need this is basically, I have a container of very complex objects, which I need to pass to certain functions that only care about one or two virtual functions in those complex objects.

我知道简短的答案是你不能， 我真的在寻找人们用来解决这个问题的任何技巧/习语。

I know the short answer is "you can't," I'm really looking for any tricks/idioms that people use to get around this problem.

提前感谢。

推荐答案

您的问题很奇怪;该主题询问如何将物品放在容器中而不丢失多态性 - 但这是乞求的问题;容器中的项目不会丢失多态性。你只需要一个基本类型的容器，一切正常。

Your question is odd; the subject asks "how do I put items in a container without losing polymorphism" - but that is begging the question; items in containers do not lose polymorphism. You just have a container of the base type and everything works.

从你的示例，它看起来你要问是如何转换一个容器的子指针到基本指针的容器？ - 答案是，你不能。子指针可以转换为基指针，子指针的容器不可以。它们是不相关的类型。虽然，请注意，shared_ptr 是可转换为shared_ptr，但只是因为他们有额外的魔法，使这项工作。容器没有这样的魔法。

From your sample, it looks what you're asking is "how do I convert a container of child pointers to a container of base pointers?" - and the answer to that is, you can't. child pointers are convertible to base pointers, containers of child pointers are not. They are unrelated types. Although, note that a shared_ptr is convertible to shared_ptr, but only because they have extra magic to make that work. The containers have no such magic.

一个答案是让chug一个模板函数（免责声明：我不是在一台带编译器的计算机上， t尝试编译这个）：

One answer would be to make chug a template function (disclaimer: I'm not on a computer with a compiler, so I haven't tried compiling this):

template<typename C, typename T>
void chug(const C<T>& container)
{
    typedef typename C<T>::iterator iter;
    for(iter i = container.begin(); i < container.end(); ++i)
    {
        (*i)->chug();
    }
}

然后chug可以取任何类型的任何容器只要它是指针的容器并且有一个chug方法。

Then chug can take any container of any type, as long as it's a container of pointers and has a chug method.

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