C ++将一个派生类转换为另一类。如何工作？

Question

我只是从基类创建了两个派生类。然后创建一个派生类的对象并转换为基类。然后从基类转换为第二个派生类，我希望它不会起作用，但确实可以。有人可以解释一下下面的代码如何工作吗？如何防止这种情况发生呢？

#include <iostream>
#include <memory>


class BaseClass
{
public:
    BaseClass() { std::cout << "This is base class\r\n";  };
    ~BaseClass() = default;
};

class DerivedClass : public BaseClass
{
public:
    DerivedClass() { std::cout << "This is Derived class\r\n"; };
    void print() { std::cout << "Print from Derived class\r\n"; };
    ~DerivedClass() = default;
};

class DerivedClass2 : public BaseClass
{
public:
    DerivedClass2() { std::cout << "This is Derived 2 class\r\n"; };
    void print123() { std::cout << "Print from Derived  2class\r\n"; };
    ~DerivedClass2() = default;
};

template<typename baseType, typename derivedType>
std::unique_ptr<derivedType> dynamicConvert(std::unique_ptr<baseType> baseObj)
{
    auto tmp = dynamic_cast<derivedType*>(baseObj.get());
    std::unique_ptr<derivedType> derivedPointer;
    if (tmp != nullptr)
    {
        baseObj.release();
        derivedPointer.reset(tmp);
    }
    return derivedPointer;
}

template<typename baseType, typename derivedType>
std::unique_ptr<derivedType> staticConvert(std::unique_ptr<baseType> baseObj)
{
    auto tmp = static_cast<derivedType*>(baseObj.get());
    std::unique_ptr<derivedType> derivedPointer;
    if (tmp != nullptr)
    {
        baseObj.release();
        derivedPointer.reset(tmp);
    }
    return derivedPointer;
}


int main()
{
    auto derived = std::make_unique<DerivedClass2>();
    auto base = dynamicConvert<DerivedClass2, BaseClass>(std::move(derived));
    auto derived1 = staticConvert<BaseClass, DerivedClass>(std::move(base));
    derived1->print();
    return 0;
}

关于，Hemant Kapoor

Answer 1

您不能，这是与static_cast签订的合同的一部分。它在编译时已解决，因此不会在运行时检查您是否输入了错误：转换为错误的类型只会触发UB。

[如果需要运行时检查的转换，请使用dynamic_cast。