pointer to 2d array

how to use a pointer to point a 2d array?
i have tried some code what is wrong in it ? can anybody explain these things in detail i always stuck in multi dimensional array and pointer things!
thanks in advance! please help me out in these things!

#include <iostream>

using namespace std;

int main()
{
    int **arry;
    int a[2][2]={{0,1},{2,3}};
    arry=a;
    cout<<arry[0][0];
    return 0;
}

closed account (1v5E3TCk)

Try this:


//Create your 2D array

int **array = new int[ 2 ];

for( int i = 0; i < 2; ++i )
{
    array[ i ]  = new int[ 2 ];
}

//Then set them to NULL

for( int i = 0; i < 2; ++i )
{
    for( int j = 0; j < 2; ++j )
    {
          array[ i ][ j ] = NULL;
    }
}

//Now you can use it

Good Luck!

hellcoder (114)

please suggest some source to read about these things in detail!

hellcoder (114)

if in declaration i use like this then what is wrong?

int *arry[2][2];
int a[2][2];
arry=a;

hellcoder (114)

@senhor
that is not working!

Cubbi (4774)

how to use a pointer to point a 2d array?

you could use typedefs to make it clearer:

#include <iostream>
int main()
{
    typedef int Array2x2[2][2];
    Array2x2 a = {{0,1}, {2,3}}; // array
    Array2x2* aptr = &a; // pointer to array
    std::cout << (*aptr)[0][0] << '\n';
}

or just write it all at once:

#include <iostream>
int main()
{
    int a[2][2] = {{0,1}, {2,3}};
    int (*aptr)[2][2] = &a;
    std::cout << (*aptr)[0][0] << '\n';
}

Your code declares int** a, which is a pointer to a pointer to an int, not a pointer to any array.

Seeing as you're trying to compile arry = a, it sounds like you're looking for a pointer to a row, which is what you get if you use a on the right side of assignment:

#include <iostream>
int main()
{
    int a[2][2] = {{0,1}, {2,3}};
    int (*arry)[2] = a; // or = &a[0];
    std::cout << arry[0][0] << '\n';
}

What's your use case anyway, why do you need this?

Last edited on

Disch (13742)

Multidimensional arrays are evil. Syntax poison.

http://www.cplusplus.com/forum/articles/17108/

That said...

1
2
3

    int **arry;
    int a[2][2]={{0,1},{2,3}};
    arry=a;

This doesn't work because a "pointer to a pointer" is not the same thing as a "2D array". A 2D array is an array of arrays. Contrary to what you might think... arrays and pointers are not the same thing.

If you want a pointer to the actual 2D array:

int a[2][2] = {...};
int (*ptr)[2][2];
ptr = &a;

cout << (*ptr)[0][0];  // accessing elements

Note this requires the size of ALL dimensions be known.
Also note this does not work with dynamically allocated ("nested new") arrays.

If you want a pointer to the first 1D array in the 2D array

int a[2][2] = {...};
int (*ptr)[2];

ptr = &a[0];
  // or
ptr = a;

cout << ptr[0][0];  // accessing elements

Note this requires the size of the 2nd dimension be known.
Also note this still does not work with 'nested new' arrays.

hellcoder (114)

in case1 why i don't need & sign ? totally confused!

#include <iostream>

using namespace std;

int main()
{
    int a[3][3]={{0,1,2},{4,5,6}};

    int (*ptr1)[3]=a;      //case 1
    cout<<(*ptr1)[2]<<endl;

    int b[3]={7,8,9};
    int (*ptr2)[3]=&b;          //case2
    cout<<(*ptr2)[2]<<endl;
    return 0;
}

Last edited on

Zhuge (4664)

The name of an array without any qualifications can be converted to a pointer. So you don't need (and don't want) the & operator.

hellcoder (114)

only a is mean to a[0] then what is wrong with
int (*ptr2)[3]=b;
here also b mean to b[0] but if i don't use & sign this won't work!

Last edited on

Zhuge (4664)

In the first case, ptr1 is an int*[], and a is int[][]. a can be converted to int*[] since the name of an array can decay to a pointer.

In the second case, ptr2 is an int*[] and b is an int[]. So while b can be converted to int*, that just points to an int and not to an array of ints like ptr2 expects. So you can use & to take the address of b (the address of an int[], which is an int*[]).

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