假设我有一个类形状和2个派生类的圆形和方形。代码是:
Shape* s1 = new circle;
现在我想将ss设为正方形,同时保留两者共有的变量。
Shape* s1 = new Square;
我该怎么做?
答案 0 :(得分:2)
通过使用引用基类的构造函数,您可以轻松复制常见的Shape
数据:
#include <assert.h>
enum class Color { red, green, blue };
class Shape {
public:
Shape() : color(red) { }
void setColor(Color new_color) { color = new_color; }
Color getColor() const { return color; }
private:
Color color;
};
class Square : public Shape {
public:
Square() { }
// Using explicit constructor to help avoid accidentally
// using the wrong type of shape.
explicit Square(const Shape &that) : Shape(that) { }
};
class Circle : public Shape {
public:
Circle() { }
explicit Circle(const Shape &that) : Shape(that) { }
};
int main(int,char**)
{
Circle circle;
circle.setColor(Color::blue);
Square square(circle);
assert(circle.getColor()==square.getColor());
}
答案 1 :(得分:1)
您可以使用复制构造函数:
Shape* s1 = new Circle;
Shape* s1 = new Square( s1 );
使用:
class Square : public Shape
{
...
public:
Square( const Circle& rhs )
{
// Copy the value you want to keep
// Respect the rules of copy constructor implementation
}
// Even better :
Square( const Shape& rhs )
{
...
}
...
};
不要忘记将圆形转换为正方形有点奇怪。
您的实施中存在内存泄漏。如果您不想使用Circle
,请将其删除。
这会更好:
Shape* s1 = new Circle;
Shape* s2 = new Square( s1 );
delete s1;
编辑:这是关于复制构造函数和分配运算符的链接:http://www.cplusplus.com/articles/y8hv0pDG/