c++ - 类在 C++ DLL 的成员函数中返回自身

Question

要在 DLL 中使用类，为了编译器之间的兼容性，来自 https://www.codeproject.com/Articles/28969/HowTo-Export-C-classes-from-a-DLL#CppMatureApproach 的方法用来。这就要求类派生自一个抽象类作为“接口(interface)”，重写抽象类的功能。如果 DLL 的类具有返回该类值的函数，您如何派生它并覆盖抽象类的函数，因为函数必须具有相同的返回值并且不能返回抽象类的值.除了制作第三个未导出的类(可以很容易地转换为将在不丢失数据的情况下导出的原始类)之外，您如何以与其他编译器一起工作的方式从 DLL 导出类？

原始代码:仅适用于相同的编译器。

#pragma once

#ifdef UNSIGNEDBIGINT_EXPORTS
#define UNSIGNEDBIGINT_API __declspec(dllexport)
#else
#define UNSIGNEDBIGINT_API __declspec(dllimport)
#endif

#include "stdafx.h"

typedef std::deque<short int> list;

//Unsigned Bigint class.



class UNSIGNEDBIGINT_API Unsigned_Bigint
{
private:
    list digits;
    Unsigned_Bigint removeIrreleventZeros(Unsigned_Bigint);

    //Arithmetic Operations
    Unsigned_Bigint add(Unsigned_Bigint);
    Unsigned_Bigint subtract(Unsigned_Bigint);
    Unsigned_Bigint multiply(Unsigned_Bigint);
    Unsigned_Bigint divide(Unsigned_Bigint, bool);

    //Comparison Operations
    bool greater(Unsigned_Bigint);

public:
    Unsigned_Bigint();
    Unsigned_Bigint(int);
    Unsigned_Bigint(list);
    Unsigned_Bigint(const Unsigned_Bigint&);
    ~Unsigned_Bigint();
    inline list getDigits() const;

    //Overloaded Arithmetic Operators
    inline Unsigned_Bigint operator+(Unsigned_Bigint);
    inline Unsigned_Bigint operator-(Unsigned_Bigint);
    inline Unsigned_Bigint operator*(Unsigned_Bigint);
    inline Unsigned_Bigint operator/(Unsigned_Bigint);
    inline Unsigned_Bigint operator%(Unsigned_Bigint);

    //Overloaded Comparison Operators
    inline bool operator==(Unsigned_Bigint);
    inline bool operator!=(Unsigned_Bigint);
    inline bool operator>(Unsigned_Bigint);
    inline bool operator<(Unsigned_Bigint);
    inline bool operator>=(Unsigned_Bigint);
    inline bool operator<=(Unsigned_Bigint);

    //Overloaded Asignment Operators
    inline void operator=(Unsigned_Bigint);
    inline void operator+=(Unsigned_Bigint);
    inline void operator-=(Unsigned_Bigint);
    inline void operator*=(Unsigned_Bigint);
    inline void operator/=(Unsigned_Bigint);
    inline void operator%=(Unsigned_Bigint);

    //Increment/Decrement
    inline Unsigned_Bigint& operator++();
    inline Unsigned_Bigint& operator--();
    inline Unsigned_Bigint operator++(int);
    inline Unsigned_Bigint operator--(int);

    //Exponent
    Unsigned_Bigint exponent(Unsigned_Bigint);

};

//Ostream
UNSIGNEDBIGINT_API inline std::ostream& operator<<(std::ostream&, Unsigned_Bigint);

编辑:还有一个问题是抽象基类把自己作为参数，因为是抽象类，所以不合法。使用引用需要对代码进行大量修改。有其他选择吗？

Answer 1

as the functions must have the same return value and the value of an abstract class can not be returned

不是真的，C++ 允许所谓的 covariant return types ，这意味着 override 方法的子类可以用子类型替换原始返回类型。但是即使你没有那个，你也可以返回一个子类的实例作为父类。不过，您可能遇到的问题是 slicing .如果您要返回值(而不是引用或指针)，那么当您使用父类接口(interface)时，子类中的其他成员将丢失。您可以使用额外的间接(例如，带有指针的结构)或其他一些策略来避免它。