如何用向量多態性隔離孩子

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();

我的問題是，在所有“針對每個”問題上，都列出了所有問題。 有沒有辦法做到這一點？ 我不想添加一個標識子類的變量。

Answer 1

在這種情況下，您需要使用dynamic_cast，因為您不能保證向量中具有派生類型。

像這樣的作品：

template <class DerivedType, class BaseType, class F>
void for_each(vector<BaseType *> elems, F && f) {
    for (auto elem : elems)
        if (auto ptr = dynamic_cast<DerivedType*>(elem))
            f(elem);
}

//usage
for_each<Wound>(allelems, [](Wound * w) { ... });
for_each<Disease>(allelems, [](Disease * d) { ... });

Answer 2

使用多態時，我傾向於在基類中使用枚舉標識符。 這樣做的好處是，您可以進行簡單的整數比較，以查看派生類型是否是該類型。 不利的一面是，如果有人想添加另一個或新的派生類型，則必須向基類枚舉注冊一個新的標識符。

您的代碼將如下所示：

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;
}