### Pointer arithmetic

Is pointer offset notation; pointer arithmetic?
Yes?
Is that a yes? I'm just checking, scared shitless is why. ( ,_,)
Yes. Also check this out:

 ``123456789101112`` ``````#include int main() { const int ia[] = {0, 1, 2, 3, 4}; const int *pi = ia; std::clog << pi[2] << '\n'; std::clog << *(pi + 2) << '\n'; std::clog << *(2 + pi) << '\n'; std::clog << 2[pi] << '\n'; }``````

What's `std::clog`?

How does `2[pi]` work?
std::clog is just like std::cout, but intended for logging.
http://cplusplus.com/reference/iostream/clog/

 How does 2[pi] work?

What do you mean?
@Catfish4:
In that context 2 is not a pointer, neither has a pointer type assigned to it.
I assume it doesn't work, but I also assume you wanted to show the op it doesn't work straight-away.

@zero117:

To make `2[pi]` work, it will become:

`((int *)(2))[pi]`

The reason why it works is:
Let's examinate this code snip:
 ``123`` ``````int Data[] = { 0, 1, 2, 3, 4 }; int * pi = Data; cout << pi[2];``````

What is written is "2".
It's the third element in the array.
Its location, in your RAM, is:
`(pi + 2)`
So, to access it you can do `pi[2]`.
But also `*(pi+2)`.

Due to mathematical rules, you will also be able to do
`*(2+pi)`

And
`((int *)(2))[pi]`
is equivalent to
`*(2+pi)`
@EssGeEich:
cl of VS2012 compiles the `2[pi]` without warning and the output is same as for the other three.

http://publib.boulder.ibm.com/infocenter/lnxpcomp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8l.doc%2Flanguage%2Fref%2Farsubex.htm
 By definition, the expression a[b] is equivalent to the expression *((a) + (b)), and, because addition is associative, it is also equivalent to b[a]. Between expressions a and b, one must be a pointer to a type T, and the other must have integral or enumeration type. The result of an array subscript is an lvalue.
Didn't know.
Learnin' every day.
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