std::advance too much?

[Gernot Frisch]

Hi,

what doi I get when I advance too much?

std::list<int> L;
L.push_back(4);
std::list<int>::iterator it = L.begin;
std::advance(it, 3);
for(; it!=L.end(); ++it)
{
// do I get here?!
}



--
-Gernot
int main(int argc, char** argv) {printf
("%silto%c%cf%cgl%ssic%ccom%c", "ma", 58, 'g', 64, "ba", 46, 10);}

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[John Harrison]

Hi,

what doi I get when I advance too much?

std::list<int> L;
L.push_back(4);
std::list<int>::iterator it = L.begin;
std::advance(it, 3);
for(; it!=L.end(); ++it)
{
// do I get here?!
}

What you get is undefined behaviour.

How about this (untested code)

template <class Iter, class Size>
void test_and_advance(Iter& curr, Iter end, Size n)
{
while (curr != end && n > 0)
{
++curr;
--n;
}
}

std::list<int>::iterator it = L.begin();
test_and_advance(it, L.end(), 3);
for(; it!=L.end(); ++it)
{

john

 

[Gernot Frisch]

What you get is undefined behaviour.

Ah. Thank you. Why doesn't this:
it = L.end(); ++it;
Make it == L.end()??

That would be obvious. Or even better - why doesn't "illegal" return
"NULL" in std::containers?

How about this (untested code)

Yes, I'll do that. I'll write my own advance function...
Thank you,
Gernot

 

[Rob Williscroft]

Gernot Frisch wrote in in comp.lang.c++:

Ah. Thank you. Why doesn't this:
it = L.end(); ++it;
Make it == L.end()??

Because it isn't in the requirements for (std) iterators that you can
call ++ on a pass-the-end iterator (which is what end() returns).

So implementing this behaviour (or any other) is useless for people
writing portable code.

That would be obvious. Or even better - why doesn't "illegal" return
"NULL" in std::containers?

std containers are for better or worse modeled on array's, and with
an array you can have a 1 past the end iterator (pointer) that can
be decremented. Having a singular (NULL) end() iterator couldn't
support that property.

Also pointers (which are iterators over array's) can't support
this beahviour.

Rob.

 

[John Harrison]

Ah. Thank you. Why doesn't this:
it = L.end(); ++it;
Make it == L.end()??

To implement that would require that an iterator know what container it is
iterating over, so that it could check the container to see if it had
reached the end. It would certainly be possible to implement iterators like
that but the standard committee decided not to require that behaviour, I
guess for efficiency reasons.

That would be obvious. Or even better - why doesn't "illegal" return
"NULL" in std::containers?

I don't understand what you mean here.

john

 

[tom_usenet]

Ah. Thank you. Why doesn't this:
it = L.end(); ++it;
Make it == L.end()??

That would be obvious. Or even better - why doesn't "illegal" return
"NULL" in std::containers?

Performance reasons. It certainly can't be done if the iterator is
actually a T*.

A good compromise is to have the poor performance in debug builds with
the checks removed for the release. See the new iterator debugging
feature from Dinkumware: www.dinkumware.com. For a free alternative,
STLport offers a debugging mode: www.stlport.com.

Tom