C++ 转换为派生类

C++ cast to derived class(C++ 转换为派生类)

问题描述

如何转换为派生类?以下方法都给出了以下错误:

How can i cast to a derived class? The below approaches all give the following error:

无法从 BaseType 转换为 DerivedType.没有构造函数可以接受源类型或构造函数重载解析不明确.

Cannot convert from BaseType to DerivedType. No constructor could take the source type, or constructor overload resolution was ambiguous.

BaseType m_baseType;

DerivedType m_derivedType = m_baseType; // gives same error

DerivedType m_derivedType = (DerivedType)m_baseType; // gives same error

DerivedType * m_derivedType = (DerivedType*) & m_baseType; // gives same error

推荐答案

这样想:

class Animal { /* Some virtual members */ };
class Dog: public Animal {};
class Cat: public Animal {};


Dog     dog;
Cat     cat;
Animal& AnimalRef1 = dog;  // Notice no cast required. (Dogs and cats are animals).
Animal& AnimalRef2 = cat;
Animal* AnimalPtr1 = &dog;
Animal* AnimlaPtr2 = &cat;

Cat&    catRef1 = dynamic_cast<Cat&>(AnimalRef1);  // Throws an exception  AnimalRef1 is a dog
Cat*    catPtr1 = dynamic_cast<Cat*>(AnimalPtr1);  // Returns NULL         AnimalPtr1 is a dog
Cat&    catRef2 = dynamic_cast<Cat&>(AnimalRef2);  // Works
Cat*    catPtr2 = dynamic_cast<Cat*>(AnimalPtr2);  // Works

// This on the other hand makes no sense
// An animal object is not a cat. Therefore it can not be treated like a Cat.
Animal  a;
Cat&    catRef1 = dynamic_cast<Cat&>(a);    // Throws an exception  Its not a CAT
Cat*    catPtr1 = dynamic_cast<Cat*>(&a);   // Returns NULL         Its not a CAT.

现在回顾你的第一句话:

Now looking back at your first statement:

Animal   animal = cat;    // This works. But it slices the cat part out and just
                          // assigns the animal part of the object.
Cat      bigCat = animal; // Makes no sense.
                          // An animal is not a cat!!!!!
Dog      bigDog = bigCat; // A cat is not a dog !!!!

您应该很少需要使用动态转换.
这就是为什么我们有虚方法:

You should very rarely ever need to use dynamic cast.
This is why we have virtual methods:

void makeNoise(Animal& animal)
{
     animal.DoNoiseMake();
}

Dog    dog;
Cat    cat;
Duck   duck;
Chicken chicken;

makeNoise(dog);
makeNoise(cat);
makeNoise(duck);
makeNoise(chicken);

我能想到的唯一原因是，如果您将对象存储在基类容器中:

The only reason I can think of is if you stored your object in a base class container:

std::vector<Animal*>  barnYard;
barnYard.push_back(&dog);
barnYard.push_back(&cat);
barnYard.push_back(&duck);
barnYard.push_back(&chicken);

Dog*  dog = dynamic_cast<Dog*>(barnYard[1]); // Note: NULL as this was the cat.

但是如果您需要将特定对象转换回 Dogs，那么您的设计就存在一个基本问题.您应该通过虚拟方法访问属性.

But if you need to cast particular objects back to Dogs then there is a fundamental problem in your design. You should be accessing properties via the virtual methods.

barnYard[1]->DoNoiseMake();

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