C++ Pointer Arithmetics

Question

Given the code:

int arr[] = {11,22,33,44,55}
for(int i = 0; i <5 ; i++)
    cout << *(arr+i) << " ";

Does *(arr+i) have the same effect as arr[i]?

Easy to test, what happens when you replace it with `arr[i]` — Barış Uşaklı, Apr 25 '13 at 19:43
@LouisTan It's nice "party knowledge", but don't ever use it in production code, it will drive anyone nuts who sees it and doesn't know it. — Christian Rau, Apr 26 '13 at 07:22

score 11 · Accepted Answer · answered Apr 25 '13 at 19:44

Yes. In fact, the subscript operator E1[E2] is defined as equivalent to *((E1)+(E2)):

A postfix expression followed by an expression in square brackets is a postfix expression. One of the expressions shall have the type “pointer to T” and the other shall have unscoped enumeration or integral type. The result is an lvalue of type “T.” The type “T” shall be a completely-defined object type. The expression E1[E2] is identical (by definition) to *((E1)+(E2)).

Which leads to something strange, like: `i[arr]`. – m0nhawk Apr 25 '13 at 19:59 — m0nhawk, Apr 25 '13 at 19:59

taocp · Answer 2 · 2013-04-25T19:58:22.933

1

yes. array are decayed to pointers. Array name points to first element of array. So

 *(arr +i)

is equivalent to:

 arr[i]

edited Apr 25 '13 at 19:58

answered Apr 25 '13 at 19:43

taocp

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C++ Pointer Arithmetics

2 Answers2