using pointers to display elements from a multidimensional array

Question

Hi Im just starting to get my head around pointers and arrays and Ive more or less know how to manipulate pointers in a one dimensional arrays to display the elements. But how about in multidimensional arrays? I have been practicing with this code:

#include<iostream>
using namespace std;
int main()
{

    int a[2][3]= { {1,2,3},{4,5,6}};
    int (*ptr)[3] = &a[0]; // or (*ptr)[3] = a;

    cout <<"Adress 0,0: "<< a << endl;
    cout <<"Adress 0,0: "<< ptr << endl;         
    cout <<"Value 0,0: "<< *a[0] << endl;
    cout <<"Value 0,0: "<< *(ptr)[0]<< endl;
    cout <<"Adress 0,1: "<< &a[0][1] << endl;
    cout <<"Adress 0,1: "<< (ptr)[1] << endl;       

    return 0;
}

I have managed to display the address and value of a[0][0] using the array name and the pointer, but how to display the address and value of a[0][1] and the succeeding elements by using the pointer?

(ptr)[1] doesn't point to a[0][1], it points to a[1][0]. — flogram_dev
– flogram_dev, Commented Feb 10, 2015 at 17:30
I think you're trying to hit (*ptr)[1] if you're looking to deliver the value 2 from your array of arrays. The address of that element is simply (*ptr)+1 (or, if you want to be long-winded, *(ptr+0) + 1 — WhozCraig
– WhozCraig, Commented Feb 10, 2015 at 17:31

flogram_dev · Accepted Answer · 2015-02-10 17:38:24Z

1

(ptr)[1] (same as ptr[1]) doesn't point to a[0][1], it points to a[1][0] because you defined ptr as a pointer to int[3], not int. Therefore incrementing ptr by 1 with ptr[1] skips three ints, up to a[1][0].

To increment ptr by the size of one int instead of three ints:

ptr[0] + 1

The above will point to a[0][1]. And to access that:

*(ptr[0] + 1)

edited Feb 10, 2015 at 17:38

answered Feb 10, 2015 at 17:25

flogram_dev

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1 Comment

datachat Over a year ago

Thank you @zenith for clarifying. Yup my previous code pointed (or skipped) 3 ints ahead thats why it printed 4 instead of 2.

Rishit Sanmukhani · Accepted Answer · 2015-02-10 17:36:41Z

1

    #include<iostream>
    using namespace std;

    int main()

{

    int a[2][3]= { {1,2,3},{4,5,6}  };
    int (*ptr)[3] = &a[0]; // or (*ptr)[3] = a;

    cout <<"Adress 0,0: "<< a << endl;
    cout <<"Adress 0,0: "<< ptr << endl;         
    cout <<"Value 0,0: "<< *a[0] << endl;
    cout <<"Value 0,0: "<< *((ptr)[0]+0)<< endl;
    cout <<"Adress 0,1: "<< &a[0][1] << endl;
    cout <<"Adress 0,1: "<< (ptr)[0]+1 << endl;       
    cout <<"value 0,1: "<<  a[0][1] << endl;
    cout <<"value 0,1: "<< *((ptr)[0]+1) << endl;

    cout <<"Adress 1,0: "<< &a[1][0] << endl;
    cout <<"Adress 1,0: "<< (ptr)[1] << endl;       
    cout <<"value 1,0: "<<  a[1][0] << endl;
    cout <<"value 1,0: "<< *((ptr)[1]+0) << endl;       

    return 0;
}

Hope this clears your doubt.

answered Feb 10, 2015 at 17:36

Rishit Sanmukhani

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1 Comment

datachat Over a year ago

Thank you for your clarification, I now understand how to traverse a 2 dimension array using pointers.

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