Accessing Array Element

[Learner010]

i learned about 1D Array , 2D Array , 3D Array.But today , suddenly my friend asked my a question on "how to access Array Elements".
i said "just use array with index".like a[0],a[1] and so on.

and then he replied(it seemed that he knew very well about array) "is it possible 0[a],1[a] and so on ?". i was confused there and tried on the compiler and run it , it run successfully , no error were there.But he could make me satisfied with the logic why the result of a[i] and i[a] are same.

i want to know the logic behind this.
why both are same ?

[deceptikon]

why both are same ?

The array index operator is syntactic sugar for an offset from a pointer. The pointer is the base address of the array, so a[0] is equivalent to *(a + 0). Because addition is commutative, you can also say *(0 + a), which falls through to the index operator and becomes 0[a].

Note, however, that while this is possible, it's extremely poor practice. Knowing about this bit of trivia is fine for impressing people at parties, but shouldn't be used in any real code.

[richieking]

As deceptikon has explained earlier,
I also wish to add something further to it.
*(0+a) is equal to calling *a or a[0].
when you add the ofset data to the pointer*(2+a) , you just ask the pointer to move to the said memory space then defference the pointer with (*) whiles the pointer has moved to the ofset.

This is called pointer arithmatic.

[Learner010]

thanks for your answers . From your replies i learn another way to access the Array Element(*(a+i) or *(i+a)). Its nice.

however i've not studied on pointer yet but i think that pointers and array are related to each other in some way.

[Tumlee]

You would be right. Arrays decay into pointers when you use them in a name context.

[Learner010]

now , my question is that how to access Elements of 2D Array using pointer ?

[deceptikon]

Exactly the same way. A 2D array "decays" to a pointer to an array, so the pointer notation is just an extension of how you do it with a 1D array:

// a[i][j]
*(*(a + i) + j)

Or equivalently:

*(a[i] + j)

Or in the opposite direction:

(*(a + i))[j]

The equivalence means these notations are interchangeable and mixable if you get your parens right.

[Learner010]

but it through an error when i use (a+i) with cin>>
i've declared and initialized as below:

a[5]={1,2,3,4,5},i;

and for the purpose to overwrite these values during runtime i wrote statement as below:

cin>>(a+i);

here error is throne. i don't know why ?

[deceptikon]

i is uninitialized in your snippets. Further, you still need to dereference in this case because operator>> is expecting a reference to the object rather than a pointer to it:

#include <algorithm>
#include <iostream>
#include <iterator>

using namespace std;

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

    cin >> *(a + i);

    copy(a, a + 5, ostream_iterator<int>(cout, " "));
    cout << '\n';
}

[richieking]

You can also use the cpp11 lambda and the for_each in the algorithm as follows.

#include <algorithm>
#include <iostream>
#include <iterator>
using namespace std;
int main()
{
    int a[] = {1, 2, 3, 4, 5};
    int i = 2;
    cin >> *(a + i);
    copy(a, a + 5, ostream_iterator<int>(cout, " "));

    // Or you can use lambda as follows
    for_each(a,a+5,[](int &c){cout<<c;});// also very useful.

    cout << '\n';
}