返回类型依赖于类的C++方法继承

Question

假设我有这门课：

class Food
{
public:
    Food * copy() const
    {
        return new Food(*this);
    }
    .
    .
    .
};

我希望它成为继承copy方法的其他类的基类。每次我都要手动写：

class Cheese: public Food
{
public:
    Cheese * copy() const
    {
        return new Cheese(*this);
    }
    .
    .
    .
};

class Bread: public Food
{
public:
    Bread * copy() const
    {
        return new Bread(*this);
    }
    .
    .
    .
};

如何解决这个问题？我在考虑模板，但我还是不知道如何使用它们.下面是我编写的程序，它不会编译：

#include <cstring>

template <class T>

class Food
{
protected:
    double weight;
public:
    T * copy() { return new T(*this); }
    Food(){ weight = 1; }
    Food(const Food& f){ weight = f.weight; }
};

class Cheese : public Food<Cheese>
{
    char * color;
public:
    Cheese(const char * c)
    {
        color = new char[strlen(c)+1];
        strcpy(color,c);
    }
    Cheese(const Cheese& c)
    {
        weight = c.weight;
        color = new char[strlen(c.color)+1];
        strcpy(color,c.color);
    }
};

main()
{
    Cheese first("yellow");
    Cheese * second = first.copy();
}

以下是一些错误：

|10|error: no matching function for call to ‘Cheese::Cheese(Food<Cheese>&)’|
|10|note: candidates are:|
|24|note: Cheese::Cheese(const Cheese&)|
|24|note:   no known conversion for argument 1 from ‘Food<Cheese>’ to ‘const Cheese&’|
|19|note: Cheese::Cheese(const char*)|
|19|note:   no known conversion for argument 1 from ‘Food<Cheese>’ to ‘const char*’|
||In member function ‘T* Food<T>::copy() [with T = Cheese]’:|
|10|warning: control reaches end of non-void function [-Wreturn-type]|

我做错了什么，最好的方法是什么？

Answer 1

搞定

T * copy() { return new T(*static_cast<T*>(this)); }

在Food<T>::copy中，'this‘是Food<T>*，而您需要一个T*。