2D数组指针-访问元素和地址

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

Answer 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) .