C++: Access via overloaded array subscript operator in 2D array

Question

For the following function that represents a matrix

class m // matrix
{
    private:

        double **matrix;
        int nrows, ncols;


        class p
        {
            private:
                double *arr;

            public:
                p (double *a)
                    :   arr (a)
                {
                }

                double &operator[] (int c)
                {
                    return arr[c];
                }
        };


    public:

        m (int nrows, int ncols)
        {
            this->matrix = new double *[nrows];
            for (int i = 0; i < nrows; ++i)
            {
                this->matrix[i] = new double [ncols];
            }
            this->nrows = nrows;
            this->ncols = ncols;
        }

        ~m()
        {
            for (int i = 0; i < this->nrows; ++i)
            {
                delete [] this->matrix[i];
            }
            delete this->matrix;
        }

        void assign (int r, int c, double v)
        {
            this->matrix[r][c] = v;
        }


        p operator[] (int r)
        {
            return p (this->matrix[r]);
        }
};

operator works for element access but does not work with element change. How can I add the functionality of assign() function to the operator?

Not yet another jagged edge matrix. Use a contiguous block of memory, pretty please. — Bathsheba
– Bathsheba, Commented Aug 26, 2015 at 7:21
possible duplicate of Overloading the subscript operator "[ ]" in the l-value and r-value cases — Amit
– Amit, Commented Aug 26, 2015 at 7:24

Bathsheba · Accepted Answer · 2015-08-26 07:23:11Z

3

Redefine your first accessor as

const p& operator[] (int r) const

and the one for setting as

p& operator[] (int r)

Then ensure that a value copy is not taken in either case. Returning a reference allows you to change the element value via the calling function.

answered Aug 26, 2015 at 7:23

Bathsheba

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2 Comments

Kudayar Pirimbaev Over a year ago

how to "ensure that a value copy is not taken in either case"? if i get it right, p that is created is destroyed after function call

Bathsheba Over a year ago

That's why you can't repeatedly create p. It always needs to be one extracted from the matrix itself.

berkus · Accepted Answer · 2015-08-26 07:30:54Z

1

Your p class is private and you won't be able to call operator [] on it.

Make p accessible, and in both operator [] implementations you should return by reference, not by value. This will allow you to modify original data, not the copy's.

answered Aug 26, 2015 at 7:30

berkus

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