Question

我想在C ++中使用迭代器，它只能迭代特定类型的元素。在以下示例中，我想仅迭代SubType实例的元素。

vector<Type*> the_vector;
the_vector.push_back(new Type(1));
the_vector.push_back(new SubType(2)); //SubType derives from Type
the_vector.push_back(new Type(3));
the_vector.push_back(new SubType(4)); 

vector<Type*>::iterator the_iterator; //***This line needs to change***

the_iterator = the_vector.begin();
while( the_iterator != the_vector.end() ) {
    SubType* item = (SubType*)*the_iterator;
    //only SubType(2) and SubType(4) should be in this loop.
    ++the_iterator;
}

如何在C ++中创建此迭代器？

Answer 1

boost filter iterator?

Answer 2

您必须使用动态广告。

the_iterator = the_vector.begin();
while( the_iterator != the_vector.end() ) {
    SubType* item = dynamic_cast<SubType*>(*the_iterator);
    if( item != 0 )
       ... 

    //only SubType(2) and SubType(4) should be in this loop.
    ++the_iterator;
}

Answer 3

没有提升的解决方案。但是，如果您可以访问boost库 - 请按照建议使用Filter Iterator。

template <typename TCollection, typename T>
class Iterator
{
public:
    typedef typename TCollection::iterator iterator;
    typedef typename TCollection::value_type value_type;

    Iterator(const TCollection& collection,
             iterator it):
        collection_(collection),
        it_(it)
    {
        moveToNextAppropriatePosition(it_);
    }
    bool operator != ( const Iterator& rhs )
    {
        return rhs.it_ != it_;
    }
    Iterator& operator++()
    {
        ++it_;
        moveToNextAppropriatePosition(it_);
        return *this;
    }
    Iterator& operator++(int);
    Iterator& operator--();
    Iterator& operator--(int);
    value_type& operator*()
    {
        return *it_;
    }
    value_type* operator->()
    {
        return &it_;
    }
private:
    const TCollection& collection_;
    iterator it_;
    void moveToNextAppropriatePosition(iterator& it)
    {
        while ( dynamic_cast<T*>(*it) == NULL && it != collection_.end() ) 
            ++it;
    }
};

class A
{
public:
    A(){}
    virtual ~A(){}
    virtual void action()
    {
        std::cout << "A";
    }
};
class B: public A
{
public:
    virtual void action()
    {
        std::cout << "B";
    }
};
int main()
{
    typedef std::vector< A* > Collection;
    Collection c;
    c.push_back( new A );
    c.push_back( new B );
    c.push_back( new A );

    typedef Iterator<Collection, B> CollectionIterator;
    CollectionIterator begin(c, c.begin());
    CollectionIterator end(c, c.end());

    std::for_each( begin, end, std::mem_fun(&A::action) );
}

Answer 4

正如paintballbob在评论中所说，你应该创建自己的迭代器类，也许继承自vector<Type*>::iterator。特别是，您需要实现或覆盖operator++()和operator++(int)以确保跳过非SubType对象（您可以使用dynamic_cast<SubType*>()检查每个项目）。在O'Reilly Net article中可以很好地概述实现自己的容器和迭代器。

Answer 5

另一种方法是如何使用boost迭代器来完成它。这一次，使用std::remove_copy_if：

std::remove_copy_if(v.begin(), v.end(), 
    boost::make_function_output_iterator(boost::bind(&someFunction, _1)),
    !boost::lambda::ll_dynamic_cast<SubType*>(boost::lambda::_1));

它将调用一个函数（在本例中为someFunction。但它可以是boost :: bind可以构造的任何东西 - 也是一个成员函数），用于指向SubType的每个指针。 / p>

如何在C ++中匹配派生类型的元素上创建迭代器？

5 个答案: