Re: index of std::vector with condition

From:

Anand Hariharan <mailto.anand.hariharan@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Tue, 13 Jul 2010 06:55:40 -0700 (PDT)

Message-ID:

<7d4041e4-062e-495a-9052-10ea7b99dd42@z34g2000pro.googlegroups.com>

On Jul 12, 8:37 pm, Stuart Golodetz
<sgolod...@NdOiSaPlA.pMiPpLeExA.ScEom> wrote:

Alf P. Steinbach /Usenet wrote:

* Stuart Golodetz, on 13.07.2010 01:53:

(...)

Anything wrong with this sort of approach?

int arr[] = {3,2,2,3,3,1,1,5,6};
std::vector<int> v(&arr[0], &arr[sizeof(arr)/sizeof(int)]);

std::map<int,std::vector<size_t> > index;
for(size_t i=0, size=v.size(); i<size; ++i)
{
index[v[i]].push_back(i);
}

Yes, it accesses non-existing elements of 'index'.

Well yes, but intentionally (and it works - I actually tried it). What
am I missing? Are you objecting on style grounds (I did lay myself wide
open <g>), or is there a more fundamental problem with the above?

Rather than -

int arr[] = {3,2,2,3,3,1,1,5,6};
std::vector<int> v(&arr[0], &arr[sizeof(arr)/sizeof(int)]);

- do

std::vector v(arr, arr + (sizeof arr / sizeof *arr));

The syntax &a[n] implies accessing the nth (or n+1 if you prefer)
element of the array and then taking its address. The typical
implementation may optimize away the dereference, but in general
&arr[sizeof(arr)/sizeof(int)] implies UB.

(And no, I do not have a C&V for this).

- Anand