Pointers pointing to Array

Question

I have written following code in C++:

#include<iostream>

using namespace std;

int main()

{
    int a[3]={2,3,1};

    int (*ptr)[3];

    ptr=&a;

    for (int i = 0; i < 3; i++)
    {
        cout<<*ptr[i]<<"            "<<ptr[i]<<"            "<<&a[i]<<endl;
    }

}

And my output is as follows:

2            0x61fe04            0x61fe04

6422020      0x61fe10            0x61fe08

2            0x61fe1c            0x61fe0c

I can understand that ptr contains the memory location of the array a as it points to the array.

But the array location(3rd col) and my pointer content (2nd col) is not a match.

and what is *ptr giving me?

Answer 1

Remember that pointers aren't arrays and arrays aren't pointers. int (*ptr)[3]; is a pointer to a int[3]. When you advance a pointer via +1 then it advances by the size of the pointee. ptr points to a int[3] hence pointer aritmethics advances in steps of sizeof(int[3]).

On the other hand a when decayed to a pointer to the first element of the array, points to an int. When you increment a it advances by sizeof(int).

Now you just need to consider that x[i] is nothing but *(x + i). Hence ptr[i] is out of bounds for any i != 0. ptr is a pointer to a int[3] but it does not point to the first element of an array of int[3].

You get expected output via:

#include<iostream>

using namespace std;

int main()

{
    int a[3]={2,3,1};

    int (*ptr)[3];

    ptr=&a;

    for (int i = 0; i < 3; i++)
    {
        cout<< (*ptr)[i]  << " " << &(*ptr)[i]<<" "<< &a[i]<<endl;
    }
}

Output:

2 0x7fff37933320 0x7fff37933320
3 0x7fff37933324 0x7fff37933324
1 0x7fff37933328 0x7fff37933328

Answer 2

Add parentheses when accessing to your pointer. * has higher priority than [], so you get incorrect results.

int main()
{
    int a[3]={2,3,1};

    int (*ptr)[3];

    ptr=&a;

    for (int i = 0; i < 3; i++)
    {
        std::cout<<(*ptr)[i]<<"            "<<&(*ptr)[i]<<"            "<<&a[i]<<std::endl;
    }
}