c – 无法调用派生类的方法 – 编译器将对象实例标识为基类
发布时间:2020-12-16 09:41:29 所属栏目:百科 来源:网络整理
导读:调用派生类中定义的方法时,我收到编译器错误.编译器似乎认为我所指的对象是基类类型: weapon = dynamic_castWeapon*(WeaponBuilder(KNIFE).name("Thief's Dagger").description("Knife favored by Thieves").attack(7) // error: class Builder has no memb
调用派生类中定义的方法时,我收到编译器错误.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()); 实际上,Builder不包含攻击或成本: 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WeaponBuilder类中找到攻击或成本方法,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. 解决方法
它找不到它,因为名称和描述都返回了Builder&而不是WeaponBuilder&,因此那些其他方法不存在.除了在任何地方进行投射之外,您的代码没有明确的解决方案.
您可以使用CRTP重写整个事物并解决您的问题,但这是一个重大变化.以下内容: 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,所有虚拟方法都应该消失,并且您将无法引用普通构建器,因为它不再是常见的基本类型,而是类模板. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |