«

int main(int argc, char* argv[])
{
std::map<int, int>::iterator footest;
{
std::map<int, int> foo;
foo[0] = 0;
footest = foo.begin();
}
return 0;
}

Is this undefined behavior? Please note that footest is going out of
scope before std::map. In theory an implementation could have a
iterator destructor that could still access the std::map ?

Thanks
Raj

Parapura Rajkumar wrote:
> int main(int argc, char* argv[])
> {
> std::map<int, int>::iterator footest;
> {
> std::map<int, int> foo;
> foo[0] = 0;
> footest = foo.begin();
> }
> return 0;
> }
>
> Is this undefined behavior? Please note that footest is going out of
> scope before std::map.

You mean, "after 'foo'", don't you?

> In theory an implementation could have a
> iterator destructor that could still access the std::map ?

No. The iterator is allowed to be invalid, that does not affect
its destruction.

V
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On Thu, 4 Jan 2007 00:47:13 -0500, "Victor Bazarov" wrote:
>Parapura Rajkumar wrote:
>> int main(int argc, char* argv[])
>> {
>> std::map<int, int>::iterator footest;
>> {
>> std::map<int, int> foo;
>> foo[0] = 0;
>> footest = foo.begin();
>> }
>> return 0;
>> }
>>
>> Is this undefined behavior? Please note that footest is going out of
>> scope before std::map.
>
>You mean, "after 'foo'", don't you?
>
>> In theory an implementation could have a
>> iterator destructor that could still access the std::map ?
>
>No. The iterator is allowed to be invalid, that does not affect
>its destruction.

Different kinds of invalid iterators are described here:
http://www.angelikalanger.com/Confer...tions-2002.pdf

Best wishes,
Roland Pibinger