c++ - 有什么方法可以让 std::unordered_set::count(Derived c)？

Question

假设我有一个抽象基类和一个具有一些独特功能的派生类:

class Shape
{
    public:
    Shape();
};

class Circle : public Shape
{
    public:
    Circle(int radius) : radius(radius);
    int getRadius() { return this->radius; };
    private:
    int radius;
};

现在我可以将 Circle*:s 放入 std::vector 中，然后检查 vector 是否具有 Shape*

std::vector< Circle* > v;

Circle* circle = new Circle(1);

v.push_back(circle);

// ...

Shape* someShape = ...; // I know it's a Circle but the pointer comes from class that only knows about Shapes

for(std::vector< Circle* >::iterator it = v.begin(); it != v.end(); ++it)
{
    Circle* c = *it;
    if((Shape)c == someShape)
    {
        // found
    }
}

// ...

delete c;

但我想使用 std::unordered_set(也适用于 std::unordered_map、std::set、std::map)，因为它具有比线性更快的 find() 和 count()

std::unordered_set< Circle* > u;

Circle* circle = new Circle(1);

u.insert(circle);

// ...

Shape* someShape = ...; // I know it's a Circle but the pointer comes from class that only knows about Shapes

if(u.count(someShape) == 1)
{
    // found
}

我得到的不是显而易见的“未定义函数”，因为只存在 std::unordered_set< Key >::count(Key& k);

有什么方法可以使用基类从 std::unordered_set 中找到 () 或 count() 吗？

Answer 1

如果您知道它是一个圆圈，那么您可以简单地将 someShape 转换为 Circle * 并将其传递给 find()。

Shape* someShape = ...; // I know it's a Circle but the pointer comes from class that only knows about Shapes
Circle* someCircle = dynamic_cast<Circle*>(someShape);

if (someCircle != NULL && u.count(someCircle) == 1)
{
    // found 
}