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如您所见,只有基类的流插入操作符的重载版本在两个实例上调用。我明白为什么如此。这是因为没有动态绑定。但是,我该如何解决它?上传和流操作符重载
#include <iostream>
using namespace std;
class A {
int i;
char c;
public:
A(int i = 0, char c = ' ') {
this->i = i;
this->c = c;
}
int getI() { return i; }
char getC() { return c; }
friend ostream& operator << (ostream&, A&);
};
class B : public A {
double d;
public:
B(int i = 0, char c = ' ', double d = 0.0) : A(i, c), d(d) {}
friend ostream& operator << (ostream&, B&);
};
ostream& operator << (ostream& out, A& a) {
out << "\nInteger: " << a.i << "\nCharacter: " << a.c << endl;
return out;
}
ostream& operator << (ostream& out, B& b) {
out << "\nInteger: " << b.getI() << "\nCharacter: " << b.getC() << "\nDouble: " << b.d << endl;
return out;
}
int main() {
A* a = new A (10, 'x');
B* b = new B(20, 'y', 5.23);
A* array[] = { a, b };
cout << *(array[0]);
cout << "\n______________________________\n";
cout << *(array[1]);
delete a;
delete b;
cin.get();
return 0;
}
我怎样才能让cout << *(array[1]);
调用重载的流插入运算符采用B的对象作为它的参数之一?
可能重复的[是否向上转型使过载流运营商(<< && >>)被一定过载作为成员函数和不友元函数?](http://stackoverflow.com/questions/29885965/does-upcasting- make-overloading-the-stream-operators-be-necessary-o) – phantom
那么解决方案在哪里? – User2k14