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When I add up in my pointer address it points to my array? Why? Reference and Dereference confusion C++

In this example, I have an array of four elements. I have declared a pointer to integer which contains the address of array. Then i have displayed the address of 0th index in 3 different ways. Similarly, the address of 1st index is displayed in 3 different ways. When I output this (&pNumbers)+0 its pretty understandable that a unique and different address would be displayed. In next line, *(&pNumbers+0) displays the 0th index address (as its contain by the pointer). Now the problem part comes. On outputting this (&pNumbers)+1 line it displays the array 0th index again. Why ?

first question: when i retrieve the pointers address (&pNumbers), it displays new and unique address. But when i add up in that address. How is it accessing the array's element addresses? e.g

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cout<<" (&pNumbers)+1 "<<(&pNumbers)+1<<endl // new address + 1 but showing address of 0 element.

cout<<" (&pNumbers + 2) "<<(&pNumbers+2)<<endl // showing address of of 1st index

second question: If say it does somehow points to the corresponding array elements. On my dereferencing why it is not displaying the correct data value corresponding to array elements. e.g 

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cout<<" *(&pNumbers+1) "<< *(&pNumbers+1) // assumption was 31 (but displaying some 0x1f)

cout<<" *(&pNumbers + 2) "<<*(&pNumbers+2)<<endl // assumption was 28 (but displpaying 0x1c)

Below is the source code: 

#include <iostream>
#include <string>
#include <conio.h>

using namespace std;

int main()
{int number[] = { 31, 28, 31, 30};
    int *pNumbers = number;

    cout<<" Address of number[0] "<<number<<endl;
    cout<<" Address of number[1] "<<number+1<<endl;
    cout<<" Address of number[2] "<<number+2<<endl;
    cout<<" Address of number[3] "<<number+3<<endl<<endl;

    cout<<" Address of &pNumbers "<< &pNumbers<<endl<<endl; // address of pNumbers

    cout<<" Address of number "<< number<<endl; // address of array's first element
    cout<<" pNumber Address "<< pNumbers<<endl;
    cout<<" &(pNumbers[0]) "<< &(pNumbers[0])<<endl<<endl;

    cout << " pNumbers+1:  " << pNumbers+1<<endl; //address of array's second element
    cout<<" (&pNumbers[1]) "<<(&pNumbers[1])<<endl; //
    cout<<" (pNumbers+1) "<< (pNumbers+1) <<endl<<endl;

    cout<<" (&pNumbers)+0 "<< (&pNumbers)+0<<endl;
    cout<<" *(&pNumbers+0) "<< *(&pNumbers+0)<<endl<<endl;

    cout<<" (&pNumbers)+1 "<<(&pNumbers)+1<<endl<<endl; // new address + 1 expected but displaying array's 0th index address why ?
    cout<<" *(&pNumbers+1) "<<*(&pNumbers+1)<<endl<<endl;

    cout<<" (&pNumbers + 2) "<<(&pNumbers+2)<<endl<<endl;
    cout<<" *(&pNumbers + 2) "<<*(&pNumbers+2)<<endl<<endl;

    cout<<" (&pNumbers + 3) "<<(&pNumbers+3)<<endl<<endl;
    cout<<" *(&pNumbers + 3) "<<*(&pNumbers+3)<<endl<<endl;

    cout<<" (&pNumbers + 4) "<<(&pNumbers+4)<<endl<<endl;
    cout<<" *(&pNumbers + 4) "<<*(&pNumbers+4)<<endl<<endl;

    return 0;
}
Solution

  • this is memory layout of your stack on most platforms. your stack

    On outputting this (&pNumbers)+1 line it displays the array 0th index again. Why ?

    Your code gets address of pNumbers memory cell (type of &pNumbers is int **), then adds one to it which semantics is to increment it by size of pointer, so expression &pNumbers + 1 is an address of your original array as pNumbers variable is a pointer itself.

    To your second question what expression *(&pNumbers+1) is doing is basically trying to interpret memory location which stores int as pointer to int (type of *(&pNumbers+1) in int *). This is the reason why you see your int in hex instead to dec format. In other words you treat number[0] as int * and print it.

    What can be confusing in all this stuff is that array is treated as pointer to its first element which causes that number and &number is essentially the same address (but both expressions has different types)

    Hope could help.