c++ 继承和共享指针

Question

情况是这样的。假设我们有一个虚拟基类(例如 ShapeJuggler)，它包含一个方法，该方法将指向虚拟基类对象(例如 Shape)的共享指针作为参数。让我们跳入以下伪代码来理解:

class Shape {
}
class ShapeJuggler {
    virtual void juggle(shared_ptr<Shape>) = 0;
}
// Now deriving a class from it
class Square : public Shape {
}
class SquareJuggler : public ShapeJuggler {
public:
    void juggle(shared_ptr<Shape>) {
         // Want to do something specific with a 'Square'
         // Or transform the 'shared_ptr<Shape>' into a 'shared_ptr<Square>'
    }
}
// Calling the juggle method
void main(void) {
    shared_ptr<Square> square_ptr = (shared_ptr<Square>) new Square();
    SquareJuggler squareJuggler;
    squareJuggler.juggle(square_ptr); // how to access 'Square'-specific members?
}

make_shared 或 dynamic/static_cast 似乎无法完成这项工作。有可能吗？有什么想法、建议吗？
谢谢

Answer 1

这是std::dynamic_pointer_cast的地方(或其 friend 之一)开始发挥作用。
它就像 dynamic_cast，但用于 std::shared_ptr。

在您的情况下(假设 Shape 类是多态的，所以 dynamic_cast 有效):

void juggle(shared_ptr<Shape> shape) {
    auto const sq = std::dynamic_pointer_cast<Square>(shape);
    assert(sq);

    sq->squareSpecificStuff();
}