c++ 2d array and pointer references

Question

In the following code when arr is passed to transpose function as below and inspect the contents of a as a[0], it is giving 0x00...001 but not the original address as inspected for arr, why it is so and what is wrong ?. I expect a[0] to be the address of 1 in the array and a[0][1] to be the first element of the array. Please explain.

problem:

int arr[][4] = { { 1, 2, 3, 4},{ 5, 6,7,8 },{ 9,10,11,12 } };
    transpose((int **)arr, 3, 4);
    int** transpose(int** a, int m, int n)
    {
        int** output = new int*[n];
        for (int i = 0;i < m;i++)
        {
            output[i] = new int[n];
        }
        for (int i = 0;i < m;i++)
        {
            for (int j = 0;j < n;j++)
            {
                //*((output[j]) + i) = *(a[i] + j);
                //*((output[j]) + i) = a[i][j];
                output[j][i] = a[i][j];
            }
        }
        return output;
    }

throwing exception.

works fine:

 int** output=transpose((int *)arr, 3, 4);
    print(output,3,4);
    int**transpose(int * a, int m, int n)
    {
        int** t = new int*[n];
        for (int i = 0;i < n;i++)
        {
            t[i] = new int[m];
        }
        for (int i = 0;i < m;i++)
        {
            for (int j = 0;j < n;j++)
            {
                t[j][i] = *((a + i*n) + j);
            }
        }
        return t;
    }

    void Matrix::print(int ** a, int m, int n)
    {
        for (int i = 0;i < m;i++)
        {
            for (int j = 0;j < n;j++)
            {
                std::cout << a[i][j] << ",";
            }
            std::cout << "\n";
        }
    }

I'm not sure you understand what "throwing exception" means. — The Quantum Physicist, Mar 27 '17 at 09:14
It is throwing acess violation exception while accessing a[i][j] in the first method. It is telling unable to read the memory. — Curious, Mar 27 '17 at 09:16
`(int **)arr` is a mistake. You seem to not understand the difference between an array and a pointer. The transpose function cannot be used with a 2-D array. — M.M, Mar 27 '17 at 09:34
yes I got to understand that 2d array are arranged serially in memory. chaning it. — Curious, Mar 27 '17 at 09:41

MNS · Accepted Answer · 2017-03-27T10:24:07.853

In order for your code to work with a 2 dimensional array, the code should be modified such as shown below.

int arr[3][4] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 } };
transpose(&arr, 3, 4);

int** transpose( int(*a)[3][4], int m, int n)
{
    int** output = new int*[n];
    for (int i = 0; i < m; i++)
    {
        output[i] = new int[n];
    }
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            //*((output[j]) + i) = *(a[i] + j);
            //*((output[j]) + i) = a[i][j];
            if (i < n && j < m )
            {
                output[j][i] = (*a)[i][j];
            }
        }
    }
    return output;

Look at the parameter declaration int(*a)[3][4]. It says that the variable ais a pointer to a 2-dimensional array of size [3][4]. An additional check if (i < n && j < m ) ensures that the array access won't go out of bounds.

It would work without any exception!

LLL · Answer 2 · 2017-03-27T09:18:58.233

0

First of all in your example m is 3 and n is 4, so in transpose function you are creating output with n (4) pointers but then you have for loop from 0 to m (3), so at this point output already has uninitialized element (output[3]). The reason why it crashes is that you use that uninitialized element in next loops.

edited Mar 27 '17 at 09:18

answered Mar 27 '17 at 09:15

LLL

1,777
1
15
31

It is throwing acess violation exception while accessing a[i][j] in the first method. It is telling unable to read the memory. – Curious Mar 27 '17 at 09:18

c++ 2d array and pointer references

2 Answers2