Path: utzoo!attcan!uunet!snorkelwacker!think!gem.mps.ohio-state.edu!mips!excelan!sjsumcs!horstman
From: horstman@sjsumcs.sjsu.edu (Cay Horstmann)
Newsgroups: comp.lang.c++
Subject: Re: operator[] for 2D array class
Summary: Double operator[] can be achieved by having the first one return X*
Message-ID: <1989Nov21.034439.27095@sjsumcs.sjsu.edu>
Date: 21 Nov 89 03:44:39 GMT
Expires: 12/15/89
References: <21162@ut-emx.UUCP>
Reply-To: horstman@sjsumcs.SJSU.EDU (Cay Horstmann)
Organization: San Jose State University
Lines: 44

In article <21162@ut-emx.UUCP> ycy@walt.cc.utexas.edu (Joseph Yip) writes:
>I am working on some routines that manipulate 2D array. In order to
>access individual elements, I define a operator [] member function.
>However, operator [] only takes one parameter. In order to do
>
>	A[i][j] = B[i][j]	// A and B are two 2D array classes
>
>I do:
>
>Tmp Array2D::operator [] (int i) {
>	...
>}
>
>int& Tmp::operator[] (int j) {
>	...
>}
>
>In this way, the problem is solved. 
>
>Is there any better way to do this object array indexing?
>

It depends. If your Array2D class internally stores all elements of a
given row consecutively somewhere, then you can have operator[] return a 
pointer to that row. Here is a naive example of a 2-dimensional array of X's

class Array2D
{	int rows, cols;
	X* elem;
	Array2D( r, c ) { rows = r; cols = c; elem = new X[r*c]; }
	X* operator[]( int n ) { if( 0 <= n && n < r ) return elem+n*cols; }
	~Array2D() { delete elem; }
};

Array2D a(3,3); 
a[1][2] = x;

a[1] returns a.elem + 1*3, a pointer to the first row of a, i.e. it skips
past the 0th row. The [2] is simply C-style [], namely *(a[1] + 2).

The trick is to return an X* so that the second [] becomes the C-style [].
Of course, this trick fails for triangular, bidiagonal or other sparse matrices.

Cay