Meaning of 1[pointer] in C++?

Question

Reading cppreference I found this example:

int a[4] = {1, 2, 3, 4};
int* p = &a[2];
std::cout << p[1] << p[-1] << 1[p] << (-1)[p] << '\n'; // 4242

I am confused about the meaning of 1[p] and (-1)[p]. Asking for help elsewhere I was told: "1[ptr] is equivalent to *(1 * sizeof(T) + ptr)". So I understand mechanically why it outputs 4242, but I find it hard to reason about. As it's nonsensical to put a pointer in square brackets usually, and is an error by itself:

[cling]$ 1[p]
(int) 4
[cling]$ [p]
input_line_15:2:3: error: 'p' cannot be captured because it does not have
      automatic storage duration
 [p]
  ^

Is this just a special case of the syntax I should just memorize, or am I misunderstanding some logic behind the statement?

There's really nothing to it aside from them choosing to make the operation commutative in that way. You don't really need to know it. It's more a fun trivia fact than something that you'll ever want to write yourself or have good reason to expect to read from anyone else's code. — Nathan Pierson, Apr 29 '22 at 02:22
`[p]` is the equivalent of `p[]` which is also invalid. Does that clear up the confusion? Have you seen: https://stackoverflow.com/questions/381542/with-arrays-why-is-it-the-case-that-a5-5a — NathanOliver, Apr 29 '22 at 02:24
Did you read point 1 of the section to which you linked? That plus the first sentence after the list explains it as well as I could. (So no need for me to waste my time if that explanation is not enough.) — JaMiT, Apr 29 '22 at 02:24
`1[pointer]` means the same as `pointer[1]`. Blame Dennis Ritchie. — user207421, Apr 29 '22 at 02:27

score 4 · Accepted Answer · answered Apr 29 '22 at 02:32

Assume p is a pointer or array, and n is an integer. When the compiler sees this:

p[n]

It logically interprets that expression as:

*(p+n)

So when it sees this:

n[p]

The compiler treats it as:

*(n+p)

Which is algebraically the same as *(p+n)

Hence: p[n] == n[p]

Meaning of 1[pointer] in C++?

1 Answers1