无法调用派生类的方法 - 编译器将对象实例标识为基类

Can't call method of derived class - compiler identifies object instance as base class

本文关键字：对象 实例 标识 基类 编译器 调用 派生 方法      更新时间：2023-10-16

当调用在派生类中定义的方法时，我得到编译器错误。编译器似乎认为我引用的对象是基类类型:

weapon = dynamic_cast<Weapon*>(WeaponBuilder(KNIFE)
.name("Thief's Dagger")
.description("Knife favored by Thieves")
.attack(7)    // error: class Builder has no member called attack 
.cost(10)     // error: class Builder has no member called cost
.build());

事实上，Builder不包含attack或cost:

class Builder
{
protected:
    string m_name;
    string m_description;
public:
    Builder();
    virtual ~Builder();
    virtual GameComponent* build() const = 0;
    Builder& name(string);
    Builder& description(string);
};

但是派生类WeaponBuilder可以:

enum WeaponType { NONE, KNIFE, SWORD, AXE, WAND };
class WeaponBuilder : public Builder
{
    int m_cost;
    int m_attack;
    int m_magic;
    WeaponType m_type;
public:
    WeaponBuilder();
    WeaponBuilder(WeaponType);
    ~WeaponBuilder();
    GameComponent* build() const;
    // should these be of reference type Builder or WeaponBuilder?
    WeaponBuilder& cost(int); 
    WeaponBuilder& attack(int);
    WeaponBuilder& magic(int);
};

我不知道为什么编译器在WeaponBuilder类中找不到attack或cost方法，因为它显然存在。我也不确定为什么它将对象识别为基类Builder的实例。

它找不到它，因为name和description都返回Builder&而不是WeaponBuilder&，因此其他方法不存在。对于你的代码没有明确的解决方案，除了到处强制转换。

您可以使用CRTP重写整个东西并解决您的问题，但这是一个重要的更改。

template< typename Derived >
class builder
{
    Derived& name( std::string const& name ){ /*store name*/, return *derived(); }
    Derived* derived(){ return static_cast< Derived* >( this ); }
};
class weapon_builder : builder< weapon_builder >
{
    weapon_builder& attack( int ){ /*store attack*/ return *this; }
    GameComponent* build() const{ return something; }
};

请注意，使用这种方法，所有virtual方法都将消失，并且您将失去引用普通builder的能力，因为它不再是公共基类型，而是类模板。

您的意图可能是这样的:

weapon = dynamic_cast<Weapon*>(dynamic_cast<WeaponBuilder &>(WeaponBuilder(KNIFE)
.name("Thief's Dagger")
.description("Knife favored by Thieves"))
.attack(7)    
.cost(10)     
.build());
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