What happens if you increment an iterator that is equal to the end iterator of an STL container

Question

What if I increment an iterator by 2 when it points onto the last element of a vector? In this question asking how to adjust the iterator to an STL container by 2 elements two different approaches are offered:

• either use a form of arithmetic operator - +=2 or ++ twice
• or use std::advance()

I've tested both of them with VC++ 7 for the edge case when the iterator points onto the last element of the STL container or beyond:

vector<int> vec;
vec.push_back( 1 );
vec.push_back( 2 );

vector<int>::iterator it = vec.begin();
advance( it, 2 );
bool isAtEnd = it == vec.end(); // true
it++; // or advance( it, 1 ); - doesn't matter
isAtEnd = it == vec.end(); //false
it = vec.begin();
advance( it, 3 );
isAtEnd = it == vec.end(); // false

I've seen may times an advise to compare against vector::end() when traversing the vector and other containers:

for( vector<int>::iterator it = vec.begin(); it != vec.end(); it++ ) {
    //manipulate the element through the iterator here
}

Obviously if the iterator is advanced past the last element inside the loop the comparison in the for-loop statement will evaluate to false and the loop will happily continue into undefined behaviour.

Do I get it right that if I ever use advance() or any kind of increment operation on an iterator and make it point past the container's end I will be unable to detect this situation? If so, what is the best practice - not to use such advancements?

Answer 1

Following is the quote from Nicolai Josuttis book:

Note that advance() does not check whether it crosses the end() of a sequence (it can't check because iterators in general do not know the containers on which they operate). Thus, calling this function might result in undefined behavior because calling operator ++ for the end of a sequence is not defined

In other words, the responsibility of maintaining the iterator within the range lies totally with the caller.