Question

我遇到以下问题：我需要使用枚举来枚举4个继承的类（此时它们与枚举之间没有区别）然后通过名为“whoAmI”的虚函数返回类型，我不明白我的语法如何会做返回部分以下是相关代码;

中的

virtual void whoAmI();
 enum gettype { easyTile, cropTile, waterTile, mediumTile};

in class.cpp

void tile::whoAmI()
{


}

Answer 1

您可以将函数的返回类型更改为enum的名称，然后使用= 0声明基类是纯虚拟的。

class ITile
{
public:
    enum class EType { easy, crop, water, medium };
    virtual EType whoAmI() const = 0;
};

然后派生类可以override此方法返回正确的enum类型，例如

class EasyTile : public ITile
{
public:
    EasyTile() = default;
    EType whoAmI() const override { return EType::easy; }
};

class CropTile : public ITile
{
public:
    CropTile() = default;
    EType whoAmI() const override { return EType::crop; }
};

以示例（live demo）

为例

int main()
{
    std::vector<std::unique_ptr<ITile>> tiles;
    tiles.emplace_back(new EasyTile);
    tiles.emplace_back(new CropTile);
    for (auto const& tile : tiles)
    {
        std::cout << static_cast<int>(tile->whoAmI()) << std::endl;
    }
}

将输出

0
1

Answer 2

您可以轻松地执行此操作：

class TileBase
{
public:
    enum Type { easyTile, cropTile, waterTile, mediumTile };

    virtual Type whoAmI() const = 0;
    virtual ~TileBase() = default;
};

class EasyTile : public TileBase
{
    Type whoAmI() const override { return easyTile; }    
};

您看，您需要将enum Type指定为返回类型而不是void。

Answer 3

#include <iostream>
using namespace std;

class Tile{
public:
    enum getType { easyTile, cropTile, waterTile, mediumTile};
    virtual getType whoAmI(){}
};
class easyTile:public Tile{
public:
    getType whoAmI(){return getType::easyTile;}
};
class cropTile: public Tile{
public:
    virtual getType whoAmI(){return getType::cropTile;}
};
class waterTile: public Tile{
public:
    virtual getType whoAmI(){return getType::waterTile;}
};
class mediumTile: public Tile{
public:
    virtual getType whoAmI(){ return getType::mediumTile;}
};

int main() {
    Tile *T = new cropTile;
    cout << T->whoAmI() << endl;
    delete T;

    return 0;
}

输出：1

枚举为具有返回值的类成员函数

3 个答案: