Re: Address one past the end of array - is this syntax a valid C++?
On Saturday, 1 February 2014 17:23:39 UTC, Peter wrote:
Assume we have an array:
int arr[5];
It's legal to refer to address arr + 5, but, of course,
illegal to refer to element arr[5] as it's not part of the
array. However, arr + n is equivalent to &arr[n].
No it's not. They're only equivalent if the expression "arr[n]"
is a valid expression.
My question
is: does this equivalence also hold for an edge case of
n = 5 (or, generally, n equal to number of elements of array)?
While there's nothing wrong with arr + 5, &arr[5] looks highly
suspicious: it looks like in the first step arr[5] is
evaluated (which introduces an undefined behaviour) which
would mean the expression as a whole is undefined.
Exactly.
Does the
equivalence still hold in this special case?
No. "arr + 5" is legal, and corresponds to a pointer one beyond
the end of the array. "&arr[5]" is undefined behavior.
C has a special rule to allow "&arr[5]". Back before C++03 (and
even before C++98, I think), there was some discussion about
allowing this in C++, but in the end, the special case was not
adopted. (I think part of the motivation for not adopting it is
that you couldn't make it work with user defined containers,
like std::vector. Something like:
std::vector<int> v(5);
int* p = &v[5];
will crash, at least in debug mode, in all of the
implementations I use.)
--
James