Question

调用派生类中定义的方法时出现编译器错误。编译器似乎认为我所指的对象是基类类型：

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);
};

但派生类WeaponBuilder确实：

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);

};

我不确定为什么编译器无法在attack类中找到cost或WeaponBuilder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。

Answer 1

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。除了在任何地方投射之外，你的代码没有明确的解决方案。

您可以使用 CRTP 重写整个内容，并解决您的问题，但这是一个重大变化。以下几行：

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，使用此方法，所有virtual方法都应该消失，并且您将无法引用普通builder，因为它不再是常见的基类型，而是类模板。

Answer 2

你的意图可能就是这样：

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

无法调用派生类的方法 - 编译器将对象实例标识为基类

2 个答案: