Question

我有一个基类：

class BaseClass {
private:
    typedef void ( BaseClass::*FunctionPtr ) ( std::string );
    FunctionPtr funcPtr;
public:
    void setFunc( FunctionPtr funcPtr ) {
        this->funcPtr = funcPtr;
    }   
}

现在我需要创建一个派生类：

class DerivativeClass: public BaseClass {
public:
    // I can overload the FunctionPtr here but for what?
    // then I also need to overload setFunc() function but I can't do it 
    // because I have too much code in a BaseClass which works with funcPtr.
    // I mean works with some function from this DerivativeClass.
    typedef void ( DerivativeClass::*FunctionPtr ) ( std::string );
    // therefore it has not any common sense in using BaseClass at all if I
    // will overload all its functions here.

    void callMe() {
        printf( "Ok!\n" );
    }
    void main() { // the program starts here
        setFunc( &DerivativeClass::callMe ); // here's my problem
    }
}

我有一个问题，因为setFunc（）仅从BaseClass获取指向函数的指针，但我需要设置一个指向DerivativeClass :: callMe函数的指针。我怎么解决这个问题？也许使用模板有一些很棒的解决方案？

Answer 1

我已经找到了解决此问题的一些很酷的解决方案。看：

template < class T >
class BaseClass {
private:
    typedef void ( T::*FunctionPtr ) ();
    FunctionPtr funcPtr;
public:
    void setFunc( FunctionPtr funcPtr ) {
        this->funcPtr = funcPtr;
    }
};

class DerivativeClass: public BaseClass < DerivativeClass > {
public:
    void callMe() {
        printf( "Ok!\n" );
    }
    void main() { // the program starts here
        setFunc( &DerivativeClass::callMe );
    }   
};

如果您对此有任何建议 - 我非常感谢：）

如何重新定义衍生类中的模式？

1 个答案: