Question

I am practicing overloading operators in C++ right now and I have a problem.
I created String class, it has just to fields one is char array other is length.
I have a String “Alice has a cat” and when I call

cout<<moj[2];

I would like to get ‘i’, but now I am getting moj + 16u adress of moj + 2 sizeof(String)
When I call

 cout<<(*moj)[2];

it works as it shoud but I would like to dereference it in overloaded operator definition. I tried many things but I can’t find solution. Please correct me.

char & operator[](int el) {return napis[el];}
const char & operator[](int el) const {return napis[el];}

AND the whole code, the important things are down the page. It’s compiling and working.

    #include <iostream>
   #include <cstdio>
   #include <stdio.h>
   #include <cstring>
  using namespace std;

 class String{
public:

//THIS IS  UNIMPORTANT------------------------------------------------------------------------------
char* napis;
int dlugosc;
   String(char* napis){
   this->napis = new char[20];
   //this->napis = napis;
   memcpy(this->napis,napis,12);
   this->dlugosc = this->length();
}

   String(const String& obiekt){
   int wrt = obiekt.dlugosc*sizeof(char);
   //cout<<"before memcpy"<<endl;
   this->napis = new char[wrt];
   memcpy(this->napis,obiekt.napis,wrt);

   //cout<<"after memcpy"<<endl;
   this->dlugosc = wrt/sizeof(char);
  }

   ~String(){
   delete[] this->napis;
   }

   int length(){
   int i = 0;
   while(napis[i] != '\0'){
       i++;
   }
   return i;
  }
        void show(){
      cout<<napis<<" dlugosc = "<<dlugosc<<endl;
 }


//THIS IS IMPORTANT
    char & operator[](int el) {return napis[el];}
    const char & operator[](int el) const {return napis[el];}
};


   int main()
   {

   String* moj = new String("Alice has a cat");
  cout<<(*moj)[2]; // IT WORKS BUI
 //  cout<<moj[2]; //I WOULD LIKE TO USE THIS ONE


   return 0;
   }

Answer 1

String* moj = new String("Alice has a cat");
cout<<(*moj)[2]; // IT WORKS BUI
//  cout<<moj[2]; //I WOULD LIKE TO USE THIS ONE

That can’t be done, the subscript operator in the later case is applied to a pointer. It is only possible to overload operators when at least one of the arguments is of user defined type (or a reference to it, but not a pointer); in this particular case the arguments are String* and 2, both fundamental types.

What you may do is drop the pointer altogether, I don’t see why you need it:

String moj("Alice has a cat");
// cout<<(*moj)[2]; <-- now this doesn't work
cout<<moj[2]; // <-- but this does