How to convert fixed size array to pointer on pointer array

Question

I`m trying to calculate determinant, but build fails. There's code, where error drops.

void getCofactor(double mat[N][N], double temp[N][N], int p, int q, int n)
    {
        int i = 0, j = 0;

        // Looping for each element of the matrix
        for (int row = 0; row < n; row++)
        {
            for (int col = 0; col < n; col++)
            {
                //  Copying into temporary matrix only those element
                //  which are not in given row and column
                if (row != p && col != q)
                {
                    temp[i][j++] = mat[row][col];

                    // Row is filled, so increase row index and
                    // reset col index
                    if (j == n - 1)
                    {
                        j = 0;
                        i++;
                    }
                }
            }
        }
    }

    double determinant(double **mat, int n)
    {
        double D = 0; // Initialize result

        //  Base case : if matrix contains single element
        if (n == 1)
            return mat[0][0];

        double temp[N][N]; // To store cofactors

        int sign = 1;  // To store sign multiplier

        // Iterate for each element of first row
        for (int f = 0; f < n; f++)
        {
            // Getting Cofactor of mat[0][f]
            getCofactor(mat, temp, 0, f, n);    //ERORRRRRRRRRRR
            D += sign * mat[0][f] * determinant(temp, n - 1);

            // terms are to be added with alternate sign
            sign = -sign;
        }

        return D;
    }

How to fix this problem? main.cpp:49:50: error: cannot convert ‘double (*)[4]’ to ‘double’ for argument ‘1’ to ‘double determinant(double**, int)’** D += sign * mat[0][f] * determinant(temp, n - 1);

Why do you pass `mat` to `determinant` as `double **` if you know it's size? use `double mat[N][N]`, or better yet, use `std::array` or `std::vector`. — Dan M., Dec 13 '18 at 13:12
You fix this problem by creating a separate array of pointers, and initializing them to point to the equivalent rows of the two-dimensional array, and then passing it. `double **` is a pointer to a pointer type. Which is not the same as an array with two dimensions. — Sam Varshavchik, Dec 13 '18 at 13:12

Ozan Yurtsever · Accepted Answer · 2018-12-13T13:28:41.267

Array with two dimensions and a pointer to another pointer are not the same things. Why do you need to use mat in different ways for two different functions such as double mat[N][N] in the function getCofactor() and double **mat in the determinant() function ? Since you know what is the value of N variable, you can use it as double mat[N][N] in the determinant() function as well.

double determinant(double mat[N][N], int n)
    {
        double D = 0; // Initialize result

        //  Base case : if matrix contains single element
        if (n == 1)
            return mat[0][0];

        double temp[N][N]; // To store cofactors

        int sign = 1;  // To store sign multiplier

        // Iterate for each element of first row
        for (int f = 0; f < n; f++)
        {
            // Getting Cofactor of mat[0][f]
            getCofactor(mat, temp, 0, f, n);    //ERORRRRRRRRRRR
            D += sign * mat[0][f] * determinant(temp, n - 1);

            // terms are to be added with alternate sign
            sign = -sign;
        }

        return D;
    }

Once you make the change I mentioned, your error will be fixed. Hope it helps.

Addition: I want to correct something which can cause a misunderstanding;

Array with two dimensions and a pointer to another pointer are not the same things.

Please note that you can have something which acts like a multidimensional array with pointers. But the way how they are occupying and storing in the RAM is not same. Note that, a multidimensional array is a single block of memory. To understand it better, check the following question;

Why can't we use double pointer to represent two dimensional arrays?

How to convert fixed size array to pointer on pointer array

1 Answers1