C++超載運算符[]

Question

如何重載operator[]使其返回b[i]，但作爲C類的對象ob2的一部分？我試圖做一個朋友methot，但它沒有奏效。

（主要任務是糾正代碼不改變主類，並使其工作）

#define MAX_CHAR 512 

class A 
{ 
protected: 
    char str[MAX_CHAR]; 
public: 
    A(char sstr[]) 
    { 
     str[0] = '\0'; 
     try 
     { 
      strcpy_s(str, sizeof(str), sstr); 
     } 
     catch (bad_alloc) 
     { } 
    } 
    void disp() 
    { 
     cout << str << endl; 
    } 
}; 

class B 
{ 
protected: 
    int* b; 
public: 
    B(int dim) 
    { 
     try 
     { 
      b = new int[dim]; 
      memset(b, 0, dim * sizeof(int)); 
     } 
     catch (bad_alloc) 
     { } 
    } 

    ~B() 
    { 
     if (b) delete[] b; 
    } 

    void disp() 
    { 
     if (b) 
      for (size_t it = 0; it < _msize(b)/sizeof(b[0]); ++it) 
       cout << b[it] << endl; 
    } 
    int& operator[](size_t dim) 
    { 
     return b[++dim]; 
    }; 
}; 


class C: public A, public B 
{ 
public: 
    C(int i, char sstr[]) 
     : A(sstr), 
      B(i) 
    { } 
    friend ostream& operator<<(ostream& wyjscie, const C& ob); 
}; 

ostream& operator<<(ostream& wyjscie, const C& ob) 
{ 
    for (int i = 0; i < 10; i++) 
     wyjscie << " i = " << *(ob.b) << " str = " << ob.str << endl; 
    return wyjscie; 
} 

int main(int argc, char* argv[]) 
{ 
    C ob1(10, "abcde"), ob2(20, "efkjyklmn"); 
    for (size_t it = 0; it < 10; ++it) 
    { 
     ob2[it] = it * it + 1; 
     cout << "ob[it] = " << ob2[it] << " it = " << it << endl; 
    } 
    cout << ob1 << endl << ob2 << endl; 
    system("pause"); 
    //ob1 = ob2; 
    //cout << endl << endl << ob1; 
    //C ob3 = ob1; 
    //cout << endl << endl << ob3; 
    return 0; 
} 

#undef MAX_CHAR 

Answer 1

class C : public A, public B 
{ 
public: 
//... 
    int operator [](int i) const 
    { 
     return b[i]; 
    } 

    int & operator [](int i) 
    { 
     return b[i]; 
    } 
}; 

或者你可以使用B類只應當正確定義已定義的操作符就像我已經顯示了C類操作員的定義一樣。