An integer literal and a string literal which gets upgraded to a memory address, yes.

The resulting memory address is implementation-dependent, platform-dependent, you name it.

It isn't a pointer, but it is an 8-bit integer (char in C terms), and it is necessarily the integer "0".

I think you mixed things up. 1 + "1" is more defined than 2 + "2", as the former points inside the string, while the latter does not.

As clarry said, don't dereference it and you're fine.

Subtract 2 from it and you have yourself a valid pointer again.

In any compiler.

So please explain why it is undefined again?

Actually, simply doing pointer arithmetic past the end of the array appears to be undefined behavior, though C++ may allow a pointer to point one past the end of the array (which might make 2 + "2" defined if you don't dereference it, but not not 3 + "3").

ADA FTW.