2d array pointers - accessing elements and address

Question

I am learning 2d array pointers and here is my code. I donot know why this line:

cout<<"Address of 1st part = "<<*ptr`

is not showing an address while this line is showing me address:

cout<<"Address of 1st part = "<<*(A)`

These both lines means same can any one help me.

#include <iostream>
using namespace std;


int main()
{
    int A[2][3]={{1,2,4},{5,8,3}};

    int *ptr;
    ptr=&A[0][0];

    cout<<"Address 1st part = "<<A<<endl;
    cout<<"Address 2nd part = "<<A+1<<endl;

    cout<<"Address 1st part = "<<ptr<<endl;
    cout<<"Address 2nd part = "<<ptr+1<<endl;

    cout<<"Address of 1st part = "<<*(A)<<endl;
    cout<<"Address of 1st part = "<<*ptr<<endl;

    cout<<"Address"<<*(A+1)+1<<endl;

    cout<<*(A+1)+2<<endl;

    return 0;
}

output

Address 1st part = 0x7fffb6c5f660 
Address 2nd part = 0x7fffb6c5f66c 
Address 1st part = 0x7fffb6c5f660 
Address 2nd part = 0x7fffb6c5f664 
Address of 1st part = 0x7fffb6c5f660 
Address of 1st part = 1 
Address0x7fffb6c5f670 
0x7fffb6c5f674

Address 1st part = 0x7fffb6c5f660 Address 2nd part = 0x7fffb6c5f66c Address 1st part = 0x7fffb6c5f660 Address 2nd part = 0x7fffb6c5f664 Address of 1st part = 0x7fffb6c5f660 Address of 1st part = 1 Address0x7fffb6c5f670 0x7fffb6c5f674 — user3215228
– user3215228, Commented Jan 26, 2014 at 20:06
not in a comment, edit your question and format it properly. — Luchian Grigore
– Luchian Grigore, Commented Jan 26, 2014 at 20:11
Because the don't mean the same thing. *ptr resolves to type int. *(A) resolves to type int (&)[3]. they're not even close. — WhozCraig
– WhozCraig, Commented Jan 26, 2014 at 20:17

Krzysztof Kosiński · Accepted Answer · 2014-01-26 20:24:41Z

1

Those two lines do not actually mean the same. A multi-dimensional array is not equivalent to a pointer to its primitive type.

A is of type int [2][3], which is equivalent to int *[3]. The type of *A is int[3], not int. The step between successive pointed-to elements, sizeof *A, is equal to sizeof(int)*3.

ptr is of type int *. The type of *ptr is int. The step here, sizeof *ptr, is equal to sizeof(int).

answered Jan 26, 2014 at 20:24

Krzysztof Kosiński

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2 Comments

user3215228 Over a year ago

Please just correct my code the line .. I just want to understand the declaration pointer variable for 2d array. I understood your point please further explain it.

Krzysztof Kosiński Over a year ago

You could write int (*ptr)[3], but I'm not entirely sure what you want to do.

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