Re: index of std::vector with condition
Anand Hariharan wrote:
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
Ah right - thanks! :)
Stu
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