Question

class Base {
public:
    virtual Base* clone() const { return new Base(*this); }
    // ...
};
class Derived: public Base {
public:
    Derived* clone() const override { return new Derived(*this); }
    // ...
};
int main() {
    Derived *d = new Derived;
    Base *b = d;
    Derived *d2 = b->clone();
    delete d;
    delete d2;
}

我在最新版本的Xcode中编译上面的代码，编译器抱怨

cannot initialize a variable of type "Derived*" with an rvalue of type "Base*"*

在Derived *d2 = b->clone()。

但我已经制作了克隆virtual并让Derived中的clone()返回Derived *。

为什么我还有这样的问题？

Answer 1

Base::clone()的返回类型为Base*，而不是Derived*。由于您通过clone()调用Base*，因此预期回报值为Base*。

如果您通过clone()调用Derived*，则可以使用Derived::clone()的返回类型。

Derived *d = new Derived;
Derived *d2 = d->clone();   // OK

此外，

Base *b = d;
Derived *d2 = dynamic_cast<Derived*>(b->clone());  // OK

此外，

Base *b = d;
Derived *d2 = dynamic_cast<Derived*>(b)->clone();  // OK

为什么编译器抱怨无法初始化＆＃34;派生＆＃34;左键型的右值

1 个答案: