class Shape { public: virtual voID draw() const {cout<<"draw shape"<<endl;}};class Point : public Shape { public: Point( int a= 0,int b = 0 ) {x=a; y=b;} // default constructor int getx() const {return x;} int gety() const {return y;} virtual voID draw() const {cout<<"draw point("<<x<<","<<y<<")\n";} private: int x,y; // x and y coordinates of Point};class Circle : public Point { public: // default constructor Circle( double r = 0.0,int x = 0,int y = 0 ):Point(x,y) {radius=r;} virtual voID draw() const {cout<<"draw circle("<<getx()<<","<<gety()<<","<<radius<<")\n";} private: double radius; // radius of Circle};voID functionCall(Shape *arrayOfShapes[3]){ Shape shape; Point point( 7,11 ); // create a Point Circle circle( 3.5,22,8 ); // create a Circle arrayOfShapes[0] = &shape; arrayOfShapes[1] = &point; arrayOfShapes[2] = &circle;}int main(){ Shape *arrayOfShapes[3]; functionCall(arrayOfShapes); for(int i=0; i<3; ++i) arrayOfShapes[i]->draw(); return 0;}
当我试图运行时,发生了分段错误.似乎主函数无法检索arrayOfShapes [3]对象?
有没有办法调用一个函数传入一个对象的指针,并在完成时返回对象的指针?
解决方法 您无法在这样的本地函数中创建形状,因为将局部变量的地址放入数组会使它们在其范围之外可用:Shape shape;arrayOfShapes[0] = &shape; // <<== Pointer to local
但是,您可以这样做:
arrayOfShapes[0] = new Shape;arrayOfShapes[1] = new Point( 7,11 );arrayOfShapes[2] = new Circle( 3.5,8 );
这种方法在动态内存中创建形状,允许在函数返回时使用它们.
注意:尽管Circle需要一个原点,但圆圈绝对不是一个点.因此,此声明在逻辑上不合理:
class Circle : public Point // This implIEs that Circle is a Point
虽然你可以说一个点是一个半径为零的圆,但是构造这样的继承也是一个坏主意.更好的方法是让Circle包含一个Point作为它的起源:
class Circle : public Shape { public: const Point origin; // default constructor Circle( double r = 0.0,int y = 0 ):origin(x,y),radius(r) {} virtual voID draw() const {cout<<"draw circle("<<origin.getx()<<","<<origin.gety()<<","<<radius<<")\n";} private: double radius; // radius of Circle};总结
以上是内存溢出为你收集整理的c – 继承多态函数调用全部内容,希望文章能够帮你解决c – 继承多态函数调用所遇到的程序开发问题。
如果觉得内存溢出网站内容还不错,欢迎将内存溢出网站推荐给程序员好友。
欢迎分享,转载请注明来源:内存溢出
评论列表(0条)