c++ - 将前向声明的指针转换为更具体的类型

Question

我有一个类

class Shape;
class Triangle;
class Amorpher
{
public:
    Amorpher();
    Amorpher(Shape*);
    Amorpher(Shape&);
    ~Amorpher();
    Shape* pShape;
    void GetShapeArea();
    void Shapeshift(Shape&, string);
    void Shapeshift(Shape*, string);
private:
    Triangle* triangle;
};

和实现

Amorpher::Amorpher()
{
}
Amorpher::Amorpher(Shape* shape) : pShape(shape){}
void Amorpher::GetShapeArea()
{
    cout << "shape area is: " << pShape->Area();

}

Amorpher::~Amorpher()
{
}
void Amorpher::Shapeshift(Shape* shape,string shiftTo)
{
    if (shiftTo == "triangle")
    {
        (Triangle*)shape = triangle;
    }
}

三角形继承自形状。我想在 Shapeshift 方法中尝试将传递给该方法的 Shape 转换为 Triangle。并非所有形状都是三角形，但为什么我不能明确地进行此转换？前向声明与问题有什么关系吗？

Answer 1

I was trying to change the type of Shape passed into the Shapeshift method to a pointer that matches the string parameter in the same function

安全转换是dynamic_cast:

dynamic_cast<Triangle*>(shape);

如果 shape 是一个 Triangle*，这将成功并且表达式的结果将是一个有效的指针。否则，它将是一个空指针。

不安全的转换为static_cast:

static_cast<Triangle*>(shape);

如果 shape 碰巧不是 Triangle*，这将是未定义的行为，但不管怎样(只要 shape 是非空的)。