Re: operator overloading

From:

"osmium" <r124c4u102@comcast.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 17 May 2007 17:25:12 -0700

Message-ID:

<5b4a39F2nhm52U1@mid.individual.net>

"ashu" writes:

Ii have studied this example of overloading [] operator, but i don`t
know exactly what happened in code .kindly explain me:

#include <iostream>
using namespace std;
const int SIZE = 3;
class atype {
int a[SIZE];
public:
atype() {
register int i;
for(i=0; i<SIZE; i++) a[i] = i;
}
int &operator[](int i) {return a[i];}

This has the same effect as
  int& bracket(int)
    {
    return a[i];
    }

where the word bracket replaces the glyphs for brackets in the original

};
int main()
{
atype ob;
cout << ob[2]; // displays 2
cout << " ";
ob[2] = 25; // [] on left of =
cout << ob[2]; // now displays 25
return 0;
}

You do realize, I hope, that the example is pointless, it is simply provided
to show you the syntax?. The constructor sets
a[k] = k for all elements in the array,
but I think you already knew that.

.rest code is ok.what happened in this line of code and how
ob[2] = 25;
what it mean? is it mean ob.a[2]=25 ? kindly explain.