What is an iterator's default value?

c++ stl default containers iterator
The Void picture The Void · Aug 3, 2010 · Viewed 52k times · Source

For any STL container that I'm using, if I declare an iterator (of this particular container type) using the iterator's default constructor, what will the iterator be initialised to?

For example, I have: 

std::list<void*> address_list;
std::list<void*>::iterator iter;

What will iter be initialised to?

Answer

UncleBens picture UncleBens · Aug 3, 2010

By convention a "NULL iterator" for containers, which is used to indicate no result, compares equal to the result of container.end(). 

 std::vector<X>::iterator iter = std::find(my_vec.begin(), my_vec.end(), x);
 if (iter == my_vec.end()) {
     //no result found; iter points to "nothing"
 }

However, since a default-constructed container iterator is not associated with any particular container, there is no good value it could take. Therefore it is just an uninitialized variable and the only legal operation to do with it is to assign a valid iterator to it.

 std::vector<X>::iterator iter;  //no particular value
 iter = some_vector.begin();  //iter is now usable

For other kinds of iterators this might not be true. E.g in case of istream_iterator, a default-constructed iterator represents (compares equal to) an istream_iterator which has reached the EOF of an input stream.

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