Question

    class problem
{
public:
virtual void show() =0;
}

class wound : public problem
{
public:
void show();
}

class disease: public problem
{
public:
void show();
}

vector<problem*> lstProb;

// I want to show all wounds only, no diseases yet
for each (wound* ouch in  lstProb)
   ouch->show();

// Here only the diseases
for each (disease* berk in  lstProb)
   berk->show();

我的问题是，在＆＃34;对于每个＆＃34;，列出了所有问题。有没有办法做到这一点？我不想添加标识子类的变量。

Answer 1

在这种情况下，您需要使用dynamic_cast，因为您无法保证向量中的派生类型。

这样的工作：

route

Answer 2

使用多态时，我倾向于在基类中使用枚举标识符。对此的专业是您可以进行简单的整数比较，以查看派生类型是否为该类型。另一方面，如果有人想要添加另一个或新的派生类型，则必须将新标识符注册到基类枚举。

您的代码看起来像这样：

class Problem {
public:
    enum Type {
        WOUND,
        DISEASE
    };  

protected:
    Type type_;

public:
    virtual void show() = 0;
    Type getType() const {
        return type_;
    } 

protected:
    explicit Problem( Type type ) : type_( type ) {}
};

class Wound : public Problem {
public:
    static unsigned counter_;
    unsigned id_;

    Wound() : Problem( Type::WOUND ) {
        counter_++;
        id_ = counter_;
    }

    void show() override {
        std::cout << "Wound " << id_ << "\n";
    }
};
unsigned Wound::counter_ = 0;


class Disease : public Problem {
public:
    static unsigned counter_;
    unsigned id_;

    Disease() : Problem( Type::DISEASE ) {
        counter_++;
        id_ = counter_;
    }

    void show() override {
        std::cout << "Disease " << id_ << "\n";
    }
};
unsigned Disease::counter_ = 0;

int main() {
    std::vector<Problem*> Probs;

    // Add 10 of each to the list: types should be alternated here
    // Vector<Problem> should look like: { wound, diesease, wound, disease...}
    for ( unsigned i = 0; i < 10; i++ ) {
        //Wound* pWound = nullptr;
        //Disease* pDisease = nullptr;

        Probs.push_back( new Wound );
        Probs.push_back( new Disease );
    }

    for (auto ouch : Probs) {
        if ( ouch->getType() == Problem::WOUND ) {
            ouch->show();
        }
    }

    std::cout << "\n";

    for (auto berk : Probs) {
        if ( berk->getType() == Problem::DISEASE ) {
            berk->show();
        }
    }

    // clean up memory
    for each (Problem* p in Probs) {
        delete p;
    }

    std::cout << "\nPress any key and enter to quit." << std::endl;
    char c;
    std::cin >> c;

    return 0;
}

如何用矢量多态性隔离孩子

2 个答案: