Why is the same value output for A[0], &A, and *A?

Question

I am doing a little experiment.

#include
#include
using namespace std;
int main()
{

  int A[5][5];
  cout<

        

Answer 1

A[0], &A, *A all are pointers of distinct types that all point to the same memory location. Same value (sort of), different types.

Expression    Symmetric                                          Type  
----------------------------------------------------------------------------
A           address of first row.                               int[5][5]
&A[0][0]    address of first element                            int* 
&A          address of 2d array                                 int(*)[5][5]   
*A         = *( A + 0) = A[0] = address of first element        int[5] = decays to int*
                                                                         in a expression

My example of 5*4 dimension char arrays:

                      A
                    +---201---202---203---204---206--+
    201             | +-----+-----+-----+-----+-----+|   
    A[0] = *(A + 0)--►| 'f' | 'o' | 'r' | 'g' | 's' ||
    207             | +-----+-----+-----+-----+-----+|
    A[1] = *(A + 1)--►| 'd' | 'o' | '\0'| '\0'| '\0'||
    213             | +-----+-----+-----+-----+-----+|
    A[2] = *(A + 2)--►| 'n' | 'o' | 't' | '\0'| '\0'||
    219             | +-----+-----+-----+-----+-----+|
    A[3] = *(A + 3)--►| 'd' | 'i' | 'e' | '\0'| '\0'||
                    | +-----+-----+-----+-----+-----+|
                    +--------------------------------+

A brief explanation about figure example.

• In figure A represents complete 2-D array starts with address 201, and
• &A gives address of complete 2-D array = 201
• *A = *(A + 0) = A[0] points to first row = 201
• Note value A[0][0] is 'f' in my example, and &A[0][0] gives address of [0][0] element = 201
• Note &A[0][0] is same as *A, because &A[0][0] => &(*(*A)) => &**A => *A

So all A[0], &A, *A, A are same but symmetrically different.

To observe difference among the A[0], &A, *A, A. Type to print sizeof() information. e.g.

 cout<

Second try to print next location address using:

 cout<<(A[0] + 1)<<" "<<(&A + 1) <<" "<<(*A + 1)<<" "<<(A + 1);

For more detailed explanation, must read this answer.