上播和流运算符重载

Upcasting and Stream Operator Overloading

本文关键字：运算符 重载      更新时间：2023-10-16

如您所见，在两个实例上仅调用基类的流插入运算符的重载版本。我明白为什么会这样。这是因为没有动态绑定。但是，我该如何解决它？

#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;
}

如何使cout << *(array[1]);调用将 B 的对象作为其参数之一的重载流插入运算符？

你不能让它调用重载运算符，因为重载是在编译时解决的。

要在运行时进行解析，即使用动态调度，您需要将执行打印的代码移动到虚拟成员函数。
然后从运算符调用它（对于基类，你只需要一个）。 

class A
{
  public:
    // ...
    // Override this in B
    virtual void print(std::ostream& o) const
    {
        o << "nInteger: " << i << "nCharacter: " << c << endl;
    }
    // ...
};
ostream& operator << (std::ostream& out, const A& a) {
    a.print(out);
    return out;
}
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