c++ - 为什么抽象类的实现看不到重载的纯虚函数？

Question

给定以下代码示例，为什么重载的 AbstractBaseClass::DoAThing( const char* ) 在实现重载的纯方法的 SomeEndClass 中作为继承方法不可见抽象 DoAThing( const char* const* ) 方法？

class AbstractBaseClass
{
    public:

        virtual void DoAThing( const char* withThis ) {}
        virtual void DoAThing( const char* const* withThat ) = 0;

        AbstractBaseClass() {}
        virtual ~AbstractBaseClass() {}
};

class SomeMiddlewareClass : public AbstractBaseClass
{
    public:

        void ThisIsCool() {}

        SomeMiddlewareClass() {}
        virtual ~SomeMiddlewareClass() {}
};

class SomeEndClass : public SomeMiddlewareClass
{
    public:

        void DoAThing( const char* const* withThat ) {}

        SomeEndClass() {}
        virtual ~SomeEndClass() {}
};

void SomeFunction()
{
    SomeEndClass* myClass = new SomeEndClass();

    myClass->DoAThing( "withThis" );
    ((SomeMiddlewareClass*)myClass)->DoAThing( "withThisToo" );

    delete myClass;
}

编译器(和索引器)在 myClass->DoAThing( "withThis"); 行上产生以下错误，而 ((SomeMiddlewareClass*)myClass)->DoAThing( "withThisToo"); 行被接受。

Invalid arguments ' Candidates are: void DoAThing(const char * const *)

no matching function for call to ‘SomeEndClass::DoAThing(const char [9])’

SomeEndClass 不应该继承 AbstractBaseClass::DoAThing( const char* ) 实现吗？我做错了什么？

Answer 1

您的 SomeEndClass::DoAThing() 函数不仅覆盖从基类继承的函数，而且隐藏基类中该函数的其他重载。

您可以将 using 声明添加到您的 SomeEndClass 类:

using SomeMiddlewareClass::DoAThing;

因此:

class SomeEndClass : public SomeMiddlewareClass
{
    public:

        using SomeMiddlewareClass::DoAThing;
//      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

        void DoAThing( const char* const* withThat ) {}

        SomeEndClass() {}
        virtual ~SomeEndClass() {}
};

通过此修复，您可以在 live example 中看到您的程序正在编译.