C ++转换为派生类 [英] C++ cast to derived class

问题描述

如何投射到派生类？以下方法都会产生以下错误：

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，或者构造函数重载解析是不明确的。

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

p>

Think like this:

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.

现在回头看第一条语句：

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 !!!!

您很少需要使用动态转换。

这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);

我可以想到的唯一原因是如果你将对象存储在一个基类容器中： p>

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.

但是如果你需要把特定的对象传回给Dogs，那么你的设计就有一个根本的问题。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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