C++:如何将派生类的容器传递给期望其基类容器的函数?

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
"; }
};

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

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，我有一个包含非常复杂对象的容器，我需要将它传递给某些函数，这些函数只关心那些复杂对象中的一两个虚函数.

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();
    }
}